Develop Reference

How can I create vertical bars at a specific position after a value? [duplicate]

Is there some easy way to pad Strings in Java?
Seems like something that should be in some StringUtil-like API, but I can't find anything that does this.

Since Java 1.5, String.format() can be used to left/right pad a given string.
public static String padRight(String s, int n) {
return String.format("%-" + n + "s", s);
}
public static String padLeft(String s, int n) {
return String.format("%" + n + "s", s);
}
...
public static void main(String args[]) throws Exception {
System.out.println(padRight("Howto", 20) + "*");
System.out.println(padLeft("Howto", 20) + "*");
}
And the output is:
Howto *
Howto*

Padding to 10 characters:
String.format("%10s", "foo").replace(' ', '*');
String.format("%-10s", "bar").replace(' ', '*');
String.format("%10s", "longer than 10 chars").replace(' ', '*');
output:
*******foo
bar*******
longer*than*10*chars
Display '*' for characters of password:
String password = "secret123";
String padded = String.format("%"+password.length()+"s", "").replace(' ', '*');
output has the same length as the password string:
secret123
*********

Apache StringUtils has several methods: leftPad, rightPad, center and repeat.
But please note that — as others have mentioned and demonstrated in this answer — String.format() and the Formatter classes in the JDK are better options. Use them over the commons code.

In Guava, this is easy:
Strings.padStart("string", 10, ' ');
Strings.padEnd("string", 10, ' ');

Something simple:
The value should be a string. convert it to string, if it's not. Like "" + 123 or Integer.toString(123)
// let's assume value holds the String we want to pad
String value = "123";
Substring start from the value length char index until end length of padded:
String padded="00000000".substring(value.length()) + value;
// now padded is "00000123"
More precise
pad right:
String padded = value + ("ABCDEFGH".substring(value.length()));
// now padded is "123DEFGH"
pad left:
String padString = "ABCDEFGH";
String padded = (padString.substring(0, padString.length() - value.length())) + value;
// now padded is "ABCDE123"

Have a look at org.apache.commons.lang.StringUtils#rightPad(String str, int size, char padChar).
But the algorithm is very simple (pad right up to size chars):
public String pad(String str, int size, char padChar)
{
StringBuilder padded = new StringBuilder(str);
while (padded.length() < size)
{
padded.append(padChar);
}
return padded.toString();
}

Besides Apache Commons, also see String.format which should be able to take care of simple padding (e.g. with spaces).

Since Java 11, String.repeat(int) can be used to left/right pad a given string.
System.out.println("*".repeat(5)+"apple");
System.out.println("apple"+"*".repeat(5));
Output:
*****apple
apple*****

public static String LPad(String str, Integer length, char car) {
return (str + String.format("%" + length + "s", "").replace(" ", String.valueOf(car))).substring(0, length);
}
public static String RPad(String str, Integer length, char car) {
return (String.format("%" + length + "s", "").replace(" ", String.valueOf(car)) + str).substring(str.length(), length + str.length());
}
LPad("Hi", 10, 'R') //gives "RRRRRRRRHi"
RPad("Hi", 10, 'R') //gives "HiRRRRRRRR"
RPad("Hi", 10, ' ') //gives "Hi "
RPad("Hi", 1, ' ') //gives "H"
//etc...

Found this on Dzone
Pad with zeros:
String.format("|%020d|", 93); // prints: |00000000000000000093|

i know this thread is kind of old and the original question was for an easy solution but if it's supposed to be really fast, you should use a char array.
public static String pad(String str, int size, char padChar)
{
if (str.length() < size)
{
char[] temp = new char[size];
int i = 0;
while (i < str.length())
{
temp[i] = str.charAt(i);
i++;
}
while (i < size)
{
temp[i] = padChar;
i++;
}
str = new String(temp);
}
return str;
}
the formatter solution is not optimal. just building the format string creates 2 new strings.
apache's solution can be improved by initializing the sb with the target size so replacing below
StringBuffer padded = new StringBuffer(str);
with
StringBuffer padded = new StringBuffer(pad);
padded.append(value);
would prevent the sb's internal buffer from growing.

This took me a little while to figure out.
The real key is to read that Formatter documentation.
// Get your data from wherever.
final byte[] data = getData();
// Get the digest engine.
final MessageDigest md5= MessageDigest.getInstance("MD5");
// Send your data through it.
md5.update(data);
// Parse the data as a positive BigInteger.
final BigInteger digest = new BigInteger(1,md5.digest());
// Pad the digest with blanks, 32 wide.
String hex = String.format(
// See: http://download.oracle.com/javase/1.5.0/docs/api/java/util/Formatter.html
// Format: %[argument_index$][flags][width]conversion
// Conversion: 'x', 'X' integral The result is formatted as a hexadecimal integer
"%1$32x",
digest
);
// Replace the blank padding with 0s.
hex = hex.replace(" ","0");
System.out.println(hex);

Here is another way to pad to the right:
// put the number of spaces, or any character you like, in your paddedString
String paddedString = "--------------------";
String myStringToBePadded = "I like donuts";
myStringToBePadded = myStringToBePadded + paddedString.substring(myStringToBePadded.length());
//result:
myStringToBePadded = "I like donuts-------";

You can reduce the per-call overhead by retaining the padding data, rather than rebuilding it every time:
public class RightPadder {
private int length;
private String padding;
public RightPadder(int length, String pad) {
this.length = length;
StringBuilder sb = new StringBuilder(pad);
while (sb.length() < length) {
sb.append(sb);
}
padding = sb.toString();
}
public String pad(String s) {
return (s.length() < length ? s + padding : s).substring(0, length);
}
}
As an alternative, you can make the result length a parameter to the pad(...) method. In that case do the adjustment of the hidden padding in that method instead of in the constructor.
(Hint: For extra credit, make it thread-safe! ;-)

All string operation usually needs to be very efficient - especially if you are working with big sets of data. I wanted something that's fast and flexible, similar to what you will get in plsql pad command. Also, I don't want to include a huge lib for just one small thing. With these considerations none of these solutions were satisfactory. This is the solutions I came up with, that had the best bench-marking results, if anybody can improve on it, please add your comment.
public static char[] lpad(char[] pStringChar, int pTotalLength, char pPad) {
if (pStringChar.length < pTotalLength) {
char[] retChar = new char[pTotalLength];
int padIdx = pTotalLength - pStringChar.length;
Arrays.fill(retChar, 0, padIdx, pPad);
System.arraycopy(pStringChar, 0, retChar, padIdx, pStringChar.length);
return retChar;
} else {
return pStringChar;
}
}
note it is called with String.toCharArray() and the result can be converted to String with new String((char[])result). The reason for this is, if you applying multiple operations you can do them all on char[] and not keep on converting between formats - behind the scenes, String is stored as char[]. If these operations were included in the String class itself, it would have been twice as efficient - speed and memory wise.

java.util.Formatter will do left and right padding. No need for odd third party dependencies (would you want to add them for something so trivial).
[I've left out the details and made this post 'community wiki' as it is not something I have a need for.]

you can use the built in StringBuilder append() and insert() methods,
for padding of variable string lengths:
AbstractStringBuilder append(CharSequence s, int start, int end) ;
For Example:
private static final String MAX_STRING = " "; //20 spaces
Set<StringBuilder> set= new HashSet<StringBuilder>();
set.add(new StringBuilder("12345678"));
set.add(new StringBuilder("123456789"));
set.add(new StringBuilder("1234567811"));
set.add(new StringBuilder("12345678123"));
set.add(new StringBuilder("1234567812234"));
set.add(new StringBuilder("1234567812222"));
set.add(new StringBuilder("12345678122334"));
for(StringBuilder padMe: set)
padMe.append(MAX_STRING, padMe.length(), MAX_STRING.length());

This works:
"".format("%1$-" + 9 + "s", "XXX").replaceAll(" ", "0")
It will fill your String XXX up to 9 Chars with a whitespace. After that all Whitespaces will be replaced with a 0. You can change the whitespace and the 0 to whatever you want...

public static String padLeft(String in, int size, char padChar) {
if (in.length() <= size) {
char[] temp = new char[size];
/* Llenado Array con el padChar*/
for(int i =0;i<size;i++){
temp[i]= padChar;
}
int posIniTemp = size-in.length();
for(int i=0;i<in.length();i++){
temp[posIniTemp]=in.charAt(i);
posIniTemp++;
}
return new String(temp);
}
return "";
}

Let's me leave an answer for some cases that you need to give left/right padding (or prefix/suffix string or spaces) before you concatenate to another string and you don't want to test length or any if condition.
The same to the selected answer, I would prefer the StringUtils of Apache Commons but using this way:
StringUtils.defaultString(StringUtils.leftPad(myString, 1))
Explain:
myString: the string I input, can be null
StringUtils.leftPad(myString, 1): if string is null, this statement would return null too
then use defaultString to give empty string to prevent concatenate null

#ck's and #Marlon Tarak's answers are the only ones to use a char[], which for applications that have several calls to padding methods per second is the best approach. However, they don't take advantage of any array manipulation optimizations and are a little overwritten for my taste; this can be done with no loops at all.
public static String pad(String source, char fill, int length, boolean right){
if(source.length() > length) return source;
char[] out = new char[length];
if(right){
System.arraycopy(source.toCharArray(), 0, out, 0, source.length());
Arrays.fill(out, source.length(), length, fill);
}else{
int sourceOffset = length - source.length();
System.arraycopy(source.toCharArray(), 0, out, sourceOffset, source.length());
Arrays.fill(out, 0, sourceOffset, fill);
}
return new String(out);
}
Simple test method:
public static void main(String... args){
System.out.println("012345678901234567890123456789");
System.out.println(pad("cats", ' ', 30, true));
System.out.println(pad("cats", ' ', 30, false));
System.out.println(pad("cats", ' ', 20, false));
System.out.println(pad("cats", '$', 30, true));
System.out.println(pad("too long for your own good, buddy", '#', 30, true));
}
Outputs:
012345678901234567890123456789
cats
cats
cats
cats$$$$$$$$$$$$$$$$$$$$$$$$$$
too long for your own good, buddy

Use this function.
private String leftPadding(String word, int length, char ch) {
return (length > word.length()) ? leftPadding(ch + word, length, ch) : word;
}
how to use?
leftPadding(month, 2, '0');
output:
01 02 03 04 .. 11 12

A lot of people have some very interesting techniques but I like to keep it simple so I go with this :
public static String padRight(String s, int n, char padding){
StringBuilder builder = new StringBuilder(s.length() + n);
builder.append(s);
for(int i = 0; i < n; i++){
builder.append(padding);
}
return builder.toString();
}
public static String padLeft(String s, int n, char padding) {
StringBuilder builder = new StringBuilder(s.length() + n);
for(int i = 0; i < n; i++){
builder.append(Character.toString(padding));
}
return builder.append(s).toString();
}
public static String pad(String s, int n, char padding){
StringBuilder pad = new StringBuilder(s.length() + n * 2);
StringBuilder value = new StringBuilder(n);
for(int i = 0; i < n; i++){
pad.append(padding);
}
return value.append(pad).append(s).append(pad).toString();
}

This is an efficient utility class for left pad, right pad, center pad and zero fill of strings in Java.
package com.example;
/**
* Utility class for left pad, right pad, center pad and zero fill.
*/
public final class StringPadding {
public static String left(String string, int length, char fill) {
if (string.length() < length) {
char[] chars = string.toCharArray();
char[] output = new char[length];
int delta = length - chars.length;
for (int i = 0; i < length; i++) {
if (i < delta) {
output[i] = fill;
} else {
output[i] = chars[i - delta];
}
}
return new String(output);
}
return string;
}
public static String right(String string, int length, char fill) {
if (string.length() < length) {
char[] chars = string.toCharArray();
char[] output = new char[length];
for (int i = 0; i < length; i++) {
if (i < chars.length) {
output[i] = chars[i];
} else {
output[i] = fill;
}
}
return new String(output);
}
return string;
}
public static String center(String string, int length, char fill) {
if (string.length() < length) {
char[] chars = string.toCharArray();
int delta = length - chars.length;
int a = (delta % 2 == 0) ? delta / 2 : delta / 2 + 1;
int b = a + chars.length;
char[] output = new char[length];
for (int i = 0; i < length; i++) {
if (i < a) {
output[i] = fill;
} else if (i < b) {
output[i] = chars[i - a];
} else {
output[i] = fill;
}
}
return new String(output);
}
return string;
}
public static String zerofill(String string, int length) {
return left(string, length, '0');
}
private StringPadding() {
}
/**
* For tests!
*/
public static void main(String[] args) {
String string = "123";
char blank = ' ';
System.out.println("left pad: [" + StringPadding.left(string, 10, blank) + "]");
System.out.println("right pad: [" + StringPadding.right(string, 10, blank) + "]");
System.out.println("center pad: [" + StringPadding.center(string, 10, blank) + "]");
System.out.println("zero fill: [" + StringPadding.zerofill(string, 10) + "]");
}
}
This is the output:
left pad: [ 123]
right pad: [123 ]
center pad: [ 123 ]
zero fill: [0000000123]

Java oneliners, no fancy library.
// 6 characters padding example
String pad = "******";
// testcases for 0, 4, 8 characters
String input = "" | "abcd" | "abcdefgh"
Pad Left, don't limit
result = pad.substring(Math.min(input.length(),pad.length())) + input;
results: "******" | "**abcd" | "abcdefgh"
Pad Right, don't limit
result = input + pad.substring(Math.min(input.length(),pad.length()));
results: "******" | "abcd**" | "abcdefgh"
Pad Left, limit to pad length
result = (pad + input).substring(input.length(), input.length() + pad.length());
results: "******" | "**abcd" | "cdefgh"
Pad Right, limit to pad length
result = (input + pad).substring(0, pad.length());
results: "******" | "abcd**" | "abcdef"

Another solution utilizing recursion.
This is compatible with all JDK versions and no external libraries are required:
private static String addPadding(final String str, final int desiredLength, final String padBy) {
String result = str;
if (str.length() >= desiredLength) {
return result;
} else {
result += padBy;
return addPadding(result, desiredLength, padBy);
}
}
NOTE: This solution will append the padding, with a little tweak you can prefix the pad value.

Here's a parallel version for those of you that have very long Strings :-)
int width = 100;
String s = "129018";
CharSequence padded = IntStream.range(0,width)
.parallel()
.map(i->i-(width-s.length()))
.map(i->i<0 ? '0' :s.charAt(i))
.collect(StringBuilder::new, (sb,c)-> sb.append((char)c), (sb1,sb2)->sb1.append(sb2));

Generalizing Eko's answer (Java 11+) a bit:
public class StringUtils {
public static String padLeft(String s, char fill, int padSize) {
if (padSize < 0) {
var err = "padSize must be >= 0 (was " + padSize + ")";
throw new java.lang.IllegalArgumentException(err);
}
int repeats = Math.max(0, padSize - s.length());
return Character.toString(fill).repeat(repeats) + s;
}
public static String padRight(String s, char fill, int padSize) {
if (padSize < 0) {
var err = "padSize must be >= 0 (was " + padSize + ")";
throw new java.lang.IllegalArgumentException(err);
}
int repeats = Math.max(0, padSize - s.length());
return s + Character.toString(fill).repeat(repeats);
}
public static void main(String[] args) {
System.out.println(padLeft("", 'x', 5)); // => xxxxx
System.out.println(padLeft("1", 'x', 5)); // => xxxx1
System.out.println(padLeft("12", 'x', 5)); // => xxx12
System.out.println(padLeft("123", 'x', 5)); // => xx123
System.out.println(padLeft("1234", 'x', 5)); // => x1234
System.out.println(padLeft("12345", 'x', 5)); // => 12345
System.out.println(padLeft("123456", 'x', 5)); // => 123456
System.out.println(padRight("", 'x', 5)); // => xxxxx
System.out.println(padRight("1", 'x', 5)); // => 1xxxx
System.out.println(padRight("12", 'x', 5)); // => 12xxx
System.out.println(padRight("123", 'x', 5)); // => 123xx
System.out.println(padRight("1234", 'x', 5)); // => 1234x
System.out.println(padRight("12345", 'x', 5)); // => 12345
System.out.println(padRight("123456", 'x', 5)); // => 123456
System.out.println(padRight("1", 'x', -1)); // => throws
}
}

For what it's worth, I was looking for something that would pad around and then I decided to code it myself. It's fairly clean and you can easily derive padLeft and padRight from this
/**
* Pads around a string, both left and right using pad as the template, aligning to the right or left as indicated.
* #param a the string to pad on both left and right
* #param pad the template to pad with, it can be of any size
* #param width the fixed width to output
* #param alignRight if true, when the input string is of odd length, adds an extra pad char to the left, so values are right aligned
* otherwise add an extra pad char to the right. When the input is of even length no extra chars will be inserted
* #return the input param a padded around.
*/
public static String padAround(String a, String pad, int width, boolean alignRight) {
if (pad.length() == 0)
throw new IllegalArgumentException("Pad cannot be an empty string!");
int delta = width - a.length();
if (delta < 1)
return a;
int half = delta / 2;
int remainder = delta % 2;
String padding = pad.repeat(((half+remainder)/pad.length()+1)); // repeating the padding to occupy all possible space
StringBuilder sb = new StringBuilder(width);
// sb.append( padding.substring(0,half + (alignRight ? 0 : remainder)));
sb.append(padding, 0, half + (alignRight ? 0 : remainder));
sb.append(a);
// sb.append( padding.substring(0,half + (alignRight ? remainder : 0)));
sb.append(padding, 0, half + (alignRight ? remainder : 0));
return sb.toString();
}
While it should be fairly fast it could prolly benefit from using a few finals here and there.

s = String to pad
n = Desired length
c = Char to pad
private String padStart( String s, int n, char c ){
if( n <= 0 ) return s;
else return String.format("%" + n + "s", s).replace(' ', c);
}

Repeat the content of a String [duplicate]

I'm looking for a simple commons method or operator that allows me to repeat some string n times. I know I could write this using a for loop, but I wish to avoid for loops whenever necessary and a simple direct method should exist somewhere.
String str = "abc";
String repeated = str.repeat(3);
repeated.equals("abcabcabc");
Related to:
repeat string javascript
Create NSString by repeating another string a given number of times
Edited
I try to avoid for loops when they are not completely necessary because:
They add to the number of lines of code even if they are tucked away in another function.
Someone reading my code has to figure out what I am doing in that for loop. Even if it is commented and has meaningful variables names, they still have to make sure it is not doing anything "clever".
Programmers love to put clever things in for loops, even if I write it to "only do what it is intended to do", that does not preclude someone coming along and adding some additional clever "fix".
They are very often easy to get wrong. For loops involving indexes tend to generate off by one bugs.
For loops often reuse the same variables, increasing the chance of really hard to find scoping bugs.
For loops increase the number of places a bug hunter has to look.

Here is the shortest version (Java 1.5+ required):
repeated = new String(new char[n]).replace("\0", s);
Where n is the number of times you want to repeat the string and s is the string to repeat.
No imports or libraries needed.

If you are using Java <= 7, this is as "concise" as it gets:
// create a string made up of n copies of string s
String.format("%0" + n + "d", 0).replace("0", s);
In Java 8 and above there is a more readable way:
// create a string made up of n copies of string s
String.join("", Collections.nCopies(n, s));
Finally, for Java 11 and above, there is a new repeat​(int count) method specifically for this purpose(link)
"abc".repeat(12);
Alternatively, if your project uses java libraries there are more options.
For Apache Commons:
StringUtils.repeat("abc", 12);
For Google Guava:
Strings.repeat("abc", 12);

String::repeat
". ".repeat(7) // Seven period-with-space pairs: . . . . . . .
New in Java 11 is the method String::repeat that does exactly what you asked for:
String str = "abc";
String repeated = str.repeat(3);
repeated.equals("abcabcabc");
Its Javadoc says:
/**
* Returns a string whose value is the concatenation of this
* string repeated {#code count} times.
* <p>
* If this string is empty or count is zero then the empty
* string is returned.
*
* #param count number of times to repeat
*
* #return A string composed of this string repeated
* {#code count} times or the empty string if this
* string is empty or count is zero
*
* #throws IllegalArgumentException if the {#code count} is
* negative.
*
* #since 11
*/

Commons Lang StringUtils.repeat()
Usage:
String str = "abc";
String repeated = StringUtils.repeat(str, 3);
repeated.equals("abcabcabc");

Java 8's String.join provides a tidy way to do this in conjunction with Collections.nCopies:
// say hello 100 times
System.out.println(String.join("", Collections.nCopies(100, "hello")));

Here's a way to do it using only standard String functions and no explicit loops:
// create a string made up of n copies of s
repeated = String.format(String.format("%%%ds", n), " ").replace(" ",s);

If you're like me and want to use Google Guava and not Apache Commons. You can use the repeat method in the Guava Strings class.
Strings.repeat("-", 60);

With java-8, you can also use Stream.generate.
import static java.util.stream.Collectors.joining;
...
String repeated = Stream.generate(() -> "abc").limit(3).collect(joining()); //"abcabcabc"
and you can wrap it in a simple utility method if needed:
public static String repeat(String str, int times) {
return Stream.generate(() -> str).limit(times).collect(joining());
}

So you want to avoid loops?
Here you have it:
public static String repeat(String s, int times) {
if (times <= 0) return "";
else return s + repeat(s, times-1);
}
(of course I know this is ugly and inefficient, but it doesn't have loops :-p)
You want it simpler and prettier? use jython:
s * 3
Edit: let's optimize it a little bit :-D
public static String repeat(String s, int times) {
if (times <= 0) return "";
else if (times % 2 == 0) return repeat(s+s, times/2);
else return s + repeat(s+s, times/2);
}
Edit2: I've done a quick and dirty benchmark for the 4 main alternatives, but I don't have time to run it several times to get the means and plot the times for several inputs... So here's the code if anybody wants to try it:
public class Repeat {
public static void main(String[] args) {
int n = Integer.parseInt(args[0]);
String s = args[1];
int l = s.length();
long start, end;
start = System.currentTimeMillis();
for (int i = 0; i < n; i++) {
if(repeatLog2(s,i).length()!=i*l) throw new RuntimeException();
}
end = System.currentTimeMillis();
System.out.println("RecLog2Concat: " + (end-start) + "ms");
start = System.currentTimeMillis();
for (int i = 0; i < n; i++) {
if(repeatR(s,i).length()!=i*l) throw new RuntimeException();
}
end = System.currentTimeMillis();
System.out.println("RecLinConcat: " + (end-start) + "ms");
start = System.currentTimeMillis();
for (int i = 0; i < n; i++) {
if(repeatIc(s,i).length()!=i*l) throw new RuntimeException();
}
end = System.currentTimeMillis();
System.out.println("IterConcat: " + (end-start) + "ms");
start = System.currentTimeMillis();
for (int i = 0; i < n; i++) {
if(repeatSb(s,i).length()!=i*l) throw new RuntimeException();
}
end = System.currentTimeMillis();
System.out.println("IterStrB: " + (end-start) + "ms");
}
public static String repeatLog2(String s, int times) {
if (times <= 0) {
return "";
}
else if (times % 2 == 0) {
return repeatLog2(s+s, times/2);
}
else {
return s + repeatLog2(s+s, times/2);
}
}
public static String repeatR(String s, int times) {
if (times <= 0) {
return "";
}
else {
return s + repeatR(s, times-1);
}
}
public static String repeatIc(String s, int times) {
String tmp = "";
for (int i = 0; i < times; i++) {
tmp += s;
}
return tmp;
}
public static String repeatSb(String s, int n) {
final StringBuilder sb = new StringBuilder();
for(int i = 0; i < n; i++) {
sb.append(s);
}
return sb.toString();
}
}
It takes 2 arguments, the first is the number of iterations (each function run with repeat times arg from 1..n) and the second is the string to repeat.
So far, a quick inspection of the times running with different inputs leaves the ranking something like this (better to worse):
Iterative StringBuilder append (1x).
Recursive concatenation log2 invocations (~3x).
Recursive concatenation linear invocations (~30x).
Iterative concatenation linear (~45x).
I wouldn't ever guessed that the recursive function was faster than the for loop :-o
Have fun(ctional xD).

This contains less characters than your question
public static String repeat(String s, int n) {
if(s == null) {
return null;
}
final StringBuilder sb = new StringBuilder(s.length() * n);
for(int i = 0; i < n; i++) {
sb.append(s);
}
return sb.toString();
}

based on fortran's answer, this is a recusive version that uses a StringBuilder:
public static void repeat(StringBuilder stringBuilder, String s, int times) {
if (times > 0) {
repeat(stringBuilder.append(s), s, times - 1);
}
}
public static String repeat(String s, int times) {
StringBuilder stringBuilder = new StringBuilder(s.length() * times);
repeat(stringBuilder, s, times);
return stringBuilder.toString();
}

using Dollar is simple as typing:
#Test
public void repeatString() {
String string = "abc";
assertThat($(string).repeat(3).toString(), is("abcabcabc"));
}
PS: repeat works also for array, List, Set, etc

I wanted a function to create a comma-delimited list of question marks for JDBC purposes, and found this post. So, I decided to take two variants and see which one performed better. After 1 million iterations, the garden-variety StringBuilder took 2 seconds (fun1), and the cryptic supposedly more optimal version (fun2) took 30 seconds. What's the point of being cryptic again?
private static String fun1(int size) {
StringBuilder sb = new StringBuilder(size * 2);
for (int i = 0; i < size; i++) {
sb.append(",?");
}
return sb.substring(1);
}
private static String fun2(int size) {
return new String(new char[size]).replaceAll("\0", ",?").substring(1);
}

OOP Solution
Nearly every answer proposes a static function as a solution, but thinking Object-Oriented (for reusability-purposes and clarity) I came up with a Solution via Delegation through the CharSequence-Interface (which also opens up usability on mutable CharSequence-Classes).
The following Class can be used either with or without Separator-String/CharSequence and each call to "toString()" builds the final repeated String.
The Input/Separator are not only limited to String-Class, but can be every Class which implements CharSequence (e.g. StringBuilder, StringBuffer, etc)!
Source-Code:
/**
* Helper-Class for Repeating Strings and other CharSequence-Implementations
* #author Maciej Schuttkowski
*/
public class RepeatingCharSequence implements CharSequence {
final int count;
CharSequence internalCharSeq = "";
CharSequence separator = "";
/**
* CONSTRUCTOR - RepeatingCharSequence
* #param input CharSequence to repeat
* #param count Repeat-Count
*/
public RepeatingCharSequence(CharSequence input, int count) {
if(count < 0)
throw new IllegalArgumentException("Can not repeat String \""+input+"\" less than 0 times! count="+count);
if(count > 0)
internalCharSeq = input;
this.count = count;
}
/**
* CONSTRUCTOR - Strings.RepeatingCharSequence
* #param input CharSequence to repeat
* #param count Repeat-Count
* #param separator Separator-Sequence to use
*/
public RepeatingCharSequence(CharSequence input, int count, CharSequence separator) {
this(input, count);
this.separator = separator;
}
#Override
public CharSequence subSequence(int start, int end) {
checkBounds(start);
checkBounds(end);
int subLen = end - start;
if (subLen < 0) {
throw new IndexOutOfBoundsException("Illegal subSequence-Length: "+subLen);
}
return (start == 0 && end == length()) ? this
: toString().substring(start, subLen);
}
#Override
public int length() {
//We return the total length of our CharSequences with the separator 1 time less than amount of repeats:
return count < 1 ? 0
: ( (internalCharSeq.length()*count) + (separator.length()*(count-1)));
}
#Override
public char charAt(int index) {
final int internalIndex = internalIndex(index);
//Delegate to Separator-CharSequence or Input-CharSequence depending on internal index:
if(internalIndex > internalCharSeq.length()-1) {
return separator.charAt(internalIndex-internalCharSeq.length());
}
return internalCharSeq.charAt(internalIndex);
}
#Override
public String toString() {
return count < 1 ? ""
: new StringBuilder(this).toString();
}
private void checkBounds(int index) {
if(index < 0 || index >= length())
throw new IndexOutOfBoundsException("Index out of Bounds: "+index);
}
private int internalIndex(int index) {
// We need to add 1 Separator-Length to total length before dividing,
// as we subtracted one Separator-Length in "length()"
return index % ((length()+separator.length())/count);
}
}
Usage-Example:
public static void main(String[] args) {
//String input = "12345";
//StringBuffer input = new StringBuffer("12345");
StringBuilder input = new StringBuilder("123");
//String separator = "<=>";
StringBuilder separator = new StringBuilder("<=");//.append('>');
int repeatCount = 2;
CharSequence repSeq = new RepeatingCharSequence(input, repeatCount, separator);
String repStr = repSeq.toString();
System.out.println("Repeat="+repeatCount+"\tSeparator="+separator+"\tInput="+input+"\tLength="+input.length());
System.out.println("CharSeq:\tLength="+repSeq.length()+"\tVal="+repSeq);
System.out.println("String :\tLength="+repStr.length()+"\tVal="+repStr);
//Here comes the Magic with a StringBuilder as Input, as you can append to the String-Builder
//and at the same Time your Repeating-Sequence's toString()-Method returns the updated String :)
input.append("ff");
System.out.println(repSeq);
//Same can be done with the Separator:
separator.append("===").append('>');
System.out.println(repSeq);
}
Example-Output:
Repeat=2 Separator=<= Input=123 Length=3
CharSeq: Length=8 Val=123<=123
String : Length=8 Val=123<=123
123ff<=123ff
123ff<====>123ff

using only JRE classes (System.arraycopy) and trying to minimize the number of temp objects you can write something like:
public static String repeat(String toRepeat, int times) {
if (toRepeat == null) {
toRepeat = "";
}
if (times < 0) {
times = 0;
}
final int length = toRepeat.length();
final int total = length * times;
final char[] src = toRepeat.toCharArray();
char[] dst = new char[total];
for (int i = 0; i < total; i += length) {
System.arraycopy(src, 0, dst, i, length);
}
return String.copyValueOf(dst);
}
EDIT
and without loops you can try with:
public static String repeat2(String toRepeat, int times) {
if (toRepeat == null) {
toRepeat = "";
}
if (times < 0) {
times = 0;
}
String[] copies = new String[times];
Arrays.fill(copies, toRepeat);
return Arrays.toString(copies).
replace("[", "").
replace("]", "").
replaceAll(", ", "");
}
EDIT 2
using Collections is even shorter:
public static String repeat3(String toRepeat, int times) {
return Collections.nCopies(times, toRepeat).
toString().
replace("[", "").
replace("]", "").
replaceAll(", ", "");
}
however I still like the first version.

Not the shortest, but (i think) the fastest way is to use the StringBuilder:
/**
* Repeat a String as many times you need.
*
* #param i - Number of Repeating the String.
* #param s - The String wich you want repeated.
* #return The string n - times.
*/
public static String repeate(int i, String s) {
StringBuilder sb = new StringBuilder();
for (int j = 0; j < i; j++)
sb.append(s);
return sb.toString();
}

If speed is your concern, then you should use as less memory copying as possible. Thus it is required to work with arrays of chars.
public static String repeatString(String what, int howmany) {
char[] pattern = what.toCharArray();
char[] res = new char[howmany * pattern.length];
int length = pattern.length;
for (int i = 0; i < howmany; i++)
System.arraycopy(pattern, 0, res, i * length, length);
return new String(res);
}
To test speed, a similar optimal method using StirngBuilder is like this:
public static String repeatStringSB(String what, int howmany) {
StringBuilder out = new StringBuilder(what.length() * howmany);
for (int i = 0; i < howmany; i++)
out.append(what);
return out.toString();
}
and the code to test it:
public static void main(String... args) {
String res;
long time;
for (int j = 0; j < 1000; j++) {
res = repeatString("123", 100000);
res = repeatStringSB("123", 100000);
}
time = System.nanoTime();
res = repeatString("123", 1000000);
time = System.nanoTime() - time;
System.out.println("elapsed repeatString: " + time);
time = System.nanoTime();
res = repeatStringSB("123", 1000000);
time = System.nanoTime() - time;
System.out.println("elapsed repeatStringSB: " + time);
}
And here the run results from my system:
elapsed repeatString: 6006571
elapsed repeatStringSB: 9064937
Note that the test for loop is to kick in JIT and have optimal results.

a straightforward one-line solution:
requires Java 8
Collections.nCopies( 3, "abc" ).stream().collect( Collectors.joining() );

for the sake of readability and portability:
public String repeat(String str, int count){
if(count <= 0) {return "";}
return new String(new char[count]).replace("\0", str);
}

If you are worried about performance, just use a StringBuilder inside the loop and do a .toString() on exit of the Loop. Heck, write your own Util Class and reuse it. 5 Lines of code max.

I really enjoy this question. There is a lot of knowledge and styles. So I can't leave it without show my rock and roll ;)
{
String string = repeat("1234567890", 4);
System.out.println(string);
System.out.println("=======");
repeatWithoutCopySample(string, 100000);
System.out.println(string);// This take time, try it without printing
System.out.println(string.length());
}
/**
* The core of the task.
*/
#SuppressWarnings("AssignmentToMethodParameter")
public static char[] repeat(char[] sample, int times) {
char[] r = new char[sample.length * times];
while (--times > -1) {
System.arraycopy(sample, 0, r, times * sample.length, sample.length);
}
return r;
}
/**
* Java classic style.
*/
public static String repeat(String sample, int times) {
return new String(repeat(sample.toCharArray(), times));
}
/**
* Java extreme memory style.
*/
#SuppressWarnings("UseSpecificCatch")
public static void repeatWithoutCopySample(String sample, int times) {
try {
Field valueStringField = String.class.getDeclaredField("value");
valueStringField.setAccessible(true);
valueStringField.set(sample, repeat((char[]) valueStringField.get(sample), times));
} catch (Exception ex) {
throw new RuntimeException(ex);
}
}
Do you like it?

public static String repeat(String str, int times) {
int length = str.length();
int size = length * times;
char[] c = new char[size];
for (int i = 0; i < size; i++) {
c[i] = str.charAt(i % length);
}
return new String(c);
}

Simple loop
public static String repeat(String string, int times) {
StringBuilder out = new StringBuilder();
while (times-- > 0) {
out.append(string);
}
return out.toString();
}

Try this out:
public static char[] myABCs = {'a', 'b', 'c'};
public static int numInput;
static Scanner in = new Scanner(System.in);
public static void main(String[] args) {
System.out.print("Enter Number of Times to repeat: ");
numInput = in.nextInt();
repeatArray(numInput);
}
public static int repeatArray(int y) {
for (int a = 0; a < y; a++) {
for (int b = 0; b < myABCs.length; b++) {
System.out.print(myABCs[b]);
}
System.out.print(" ");
}
return y;
}

Using recursion, you can do the following (using ternary operators, one line max):
public static final String repeat(String string, long number) {
return number == 1 ? string : (number % 2 == 0 ? repeat(string + string, number / 2) : string + repeat(string + string, (number - 1) / 2));
}
I know, it's ugly and probably not efficient, but it's one line!

If you only know the length of the output string (and it may be not divisible by the length of the input string), then use this method:
static String repeat(String s, int length) {
return s.length() >= length ? s.substring(0, length) : repeat(s + s, length);
}
Usage demo:
for (int i = 0; i < 50; i++)
System.out.println(repeat("_/‾\\", i));
Don't use with empty s and length > 0, since it's impossible to get the desired result in this case.

Despite your desire not to use loops, I think you should use a loop.
String repeatString(String s, int repetitions)
{
if(repetitions < 0) throw SomeException();
else if(s == null) return null;
StringBuilder stringBuilder = new StringBuilder(s.length() * repetitions);
for(int i = 0; i < repetitions; i++)
stringBuilder.append(s);
return stringBuilder.toString();
}
Your reasons for not using a for loop are not good ones. In response to your criticisms:
Whatever solution you use will almost certainly be longer than this. Using a pre-built function only tucks it under more covers.
Someone reading your code will have to figure out what you're doing in that non-for-loop. Given that a for-loop is the idiomatic way to do this, it would be much easier to figure out if you did it with a for loop.
Yes someone might add something clever, but by avoiding a for loop you are doing something clever. That's like shooting yourself in the foot intentionally to avoid shooting yourself in the foot by accident.
Off-by-one errors are also mind-numbingly easy to catch with a single test. Given that you should be testing your code, an off-by-one error should be easy to fix and catch. And it's worth noting: the code above does not contain an off-by-one error. For loops are equally easy to get right.
So don't reuse variables. That's not the for-loop's fault.
Again, so does whatever solution you use. And as I noted before; a bug hunter will probably be expecting you to do this with a for loop, so they'll have an easier time finding it if you use a for loop.

here is the latest Stringutils.java StringUtils.java
public static String repeat(String str, int repeat) {
// Performance tuned for 2.0 (JDK1.4)
if (str == null) {
return null;
}
if (repeat <= 0) {
return EMPTY;
}
int inputLength = str.length();
if (repeat == 1 || inputLength == 0) {
return str;
}
if (inputLength == 1 && repeat <= PAD_LIMIT) {
return repeat(str.charAt(0), repeat);
}
int outputLength = inputLength * repeat;
switch (inputLength) {
case 1 :
return repeat(str.charAt(0), repeat);
case 2 :
char ch0 = str.charAt(0);
char ch1 = str.charAt(1);
char[] output2 = new char[outputLength];
for (int i = repeat * 2 - 2; i >= 0; i--, i--) {
output2[i] = ch0;
output2[i + 1] = ch1;
}
return new String(output2);
default :
StringBuilder buf = new StringBuilder(outputLength);
for (int i = 0; i < repeat; i++) {
buf.append(str);
}
return buf.toString();
}
}
it doesn't even need to be this big, can be made into this, and can be copied and pasted
into a utility class in your project.
public static String repeat(String str, int num) {
int len = num * str.length();
StringBuilder sb = new StringBuilder(len);
for (int i = 0; i < times; i++) {
sb.append(str);
}
return sb.toString();
}
So e5, I think the best way to do this would be to simply use the above mentioned code,or any of the answers here. but commons lang is just too big if it's a small project

I created a recursive method that do the same thing you want.. feel free to use this...
public String repeat(String str, int count) {
return count > 0 ? repeat(str, count -1) + str: "";
}
i have the same answer on Can I multiply strings in java to repeat sequences?

public static String rep(int a,String k)
{
if(a<=0)
return "";
else
{a--;
return k+rep(a,k);
}
You can use this recursive method for you desired goal.

How to split numbers in java?

I am trying to split number but don't know how to do this.
I referred this link.
https://stackoverflow.com/a/12297231/1395259
Suppose I have a number 12345678.
using above link I am splitting by 3 places.So the output becomes 123 456 78.
But I want it as 12 345 678 And I want to take the string that was split in the form 12.345.678 .
Can anybody help me please??

java.text package provides all reasonable options for formatting numbers
DecimalFormat f = new DecimalFormat("#,###");
DecimalFormatSymbols fs = f.getDecimalFormatSymbols();
fs.setGroupingSeparator('.');
f.setDecimalFormatSymbols(fs);
String s = f.format(12345678);
System.out.println(s);
output
12.345.678
using DecimalFormat directly is very flexible, but typically we can use a shorter version
String s = NumberFormat.getNumberInstance(Locale.GERMAN).format(12345678);
which produces the same string
12.345.678
Different countries have different rules for formatting numbers (and dates). Usually we want our programs to be internationalized / localized and use default locale version
NumberFormat.getNumberInstance().format(number);

One lazy way is to reverse the string, apply above method, and then reverse it again.
You can use StringBuffer's Reverse Function, as shown in Reverse each individual word of "Hello World" string with Java
12345678
87654321
876 543 21
12 345 678
I am assuming of course that you want to split by 3s and the group with <3 digits left appears in the start rather than the end as in the method you link.
The not so lazy way would be to use string length to adapt the method you link to start with length%3 characters.

Using the solution from your link i would rephrase that as following (TESTED!):
public String[] splitStringEvery(String s, int interval) {
int arrayLength = (int) Math.ceil(((s.length() / (double)interval)));
String[] result = new String[arrayLength];
int j = s.length();
int lastIndex = result.length;
for (int i = lastIndex - 1; i > 0; i--) {
result[i] = s.substring(j - interval, j);
j -= interval;
} //Add the last bit
result[0] = s.substring(0, j);
return result;
}

Here is a method that splits an int value and returns an String in the specified format:
public static String split( int n ) {
String result = "", s = Integer.toString( n );
while ( s.length() > 3 ) {
result = s.substring( s.length() -3, s.length() ) + ((result.length() > 0)? "." + result : "" );
s = s.substring( 0, s.length() -3 );
}
return s + "." + result;
}
Input:
12345678
Output:
12.345.678

If it's a String, use StringBuilder or StringBuffer. Here's the code:
public class SplitNumber {
public static void main(String[] args){
int number = 12345678;
String numberStrBefore = Integer.toString(number);
StringBuffer numberStrAfter = new StringBuffer();
numberStrAfter.append(numberStrBefore.charAt(0));
numberStrAfter.append(numberStrBefore.charAt(1));
numberStrAfter.append('.');
numberStrAfter.append(numberStrBefore.charAt(2));
numberStrAfter.append(numberStrBefore.charAt(3));
numberStrAfter.append(numberStrBefore.charAt(4));
numberStrAfter.append('.');
numberStrAfter.append(numberStrBefore.charAt(5));
numberStrAfter.append(numberStrBefore.charAt(6));
numberStrAfter.append(numberStrBefore.charAt(7));
System.out.println("Number Before: " + numberStrBefore);
System.out.println("Number After: " + numberStrAfter.toString());
}
}
And here is the same thing with a method:
public class SplitNumber {
public static void main(String[] args){
int number = 12345678;
int[] split = {2,3,3}; //How to split the number
String numberStrAfter = insertDots(number, split);
System.out.println("Number Before: " + number);
System.out.println("Number After: " + numberStrAfter);
}
public static String insertDots(int number, int[] split){
StringBuffer numberStrAfter = new StringBuffer();
String numberStr = Integer.toString(number);
int currentIndex = 0;
for(int i = 0; i < split.length; i++){
for(int j = 0; j < split[i]; j++){
numberStrAfter.append(numberStr.charAt(currentIndex));
currentIndex++;
}
numberStrAfter.append('.');
}
numberStrAfter.deleteCharAt(numberStrAfter.length()-1); //Removing last "."
return numberStrAfter.toString();
}
}
This version, with the method, allows you to split any number into any format that you want, simply change the "split" variable into the format that you want to split the string into. (Ex: Splitting 12345678 into: 1.1234.5.67.8 would mean that "split" must be set to {1,4,1,2,1}).

How to remove single character from a String by index

For accessing individual characters of a String in Java, we have String.charAt(2). Is there any inbuilt function to remove an individual character of a String in java?
Something like this:
if(String.charAt(1) == String.charAt(2){
//I want to remove the individual character at index 2.
}

You can also use the StringBuilder class which is mutable.
StringBuilder sb = new StringBuilder(inputString);
It has the method deleteCharAt(), along with many other mutator methods.
Just delete the characters that you need to delete and then get the result as follows:
String resultString = sb.toString();
This avoids creation of unnecessary string objects.

You can use Java String method called replace, which will replace all characters matching the first parameter with the second parameter:
String a = "Cool";
a = a.replace("o","");

One possibility:
String result = str.substring(0, index) + str.substring(index+1);
Note that the result is a new String (as well as two intermediate String objects), because Strings in Java are immutable.

No, because Strings in Java are immutable. You'll have to create a new string removing the character you don't want.
For replacing a single char c at index position idx in string str, do something like this, and remember that a new string will be created:
String newstr = str.substring(0, idx) + str.substring(idx + 1);

String str = "M1y java8 Progr5am";
deleteCharAt()
StringBuilder build = new StringBuilder(str);
System.out.println("Pre Builder : " + build);
build.deleteCharAt(1); // Shift the positions front.
build.deleteCharAt(8-1);
build.deleteCharAt(15-2);
System.out.println("Post Builder : " + build);
replace()
StringBuffer buffer = new StringBuffer(str);
buffer.replace(1, 2, ""); // Shift the positions front.
buffer.replace(7, 8, "");
buffer.replace(13, 14, "");
System.out.println("Buffer : "+buffer);
char[]
char[] c = str.toCharArray();
String new_Str = "";
for (int i = 0; i < c.length; i++) {
if (!(i == 1 || i == 8 || i == 15))
new_Str += c[i];
}
System.out.println("Char Array : "+new_Str);

To modify Strings, read about StringBuilder because it is mutable except for immutable String. Different operations can be found here https://docs.oracle.com/javase/tutorial/java/data/buffers.html. The code snippet below creates a StringBuilder and then append the given String and then delete the first character from the String and then convert it back from StringBuilder to a String.
StringBuilder sb = new StringBuilder();
sb.append(str);
sb.deleteCharAt(0);
str = sb.toString();

Consider the following code:
public String removeChar(String str, Integer n) {
String front = str.substring(0, n);
String back = str.substring(n+1, str.length());
return front + back;
}

You may also use the (huge) regexp machine.
inputString = inputString.replaceFirst("(?s)(.{2}).(.*)", "$1$2");
"(?s)" - tells regexp to handle newlines like normal characters (just in case).
"(.{2})" - group $1 collecting exactly 2 characters
"." - any character at index 2 (to be squeezed out).
"(.*)" - group $2 which collects the rest of the inputString.
"$1$2" - putting group $1 and group $2 together.

If you want to remove a char from a String str at a specific int index:
public static String removeCharAt(String str, int index) {
// The part of the String before the index:
String str1 = str.substring(0,index);
// The part of the String after the index:
String str2 = str.substring(index+1,str.length());
// These two parts together gives the String without the specified index
return str1+str2;
}

By the using replace method we can change single character of string.
string= string.replace("*", "");

Use replaceFirst function of String class. There are so many variants of replace function that you can use.

If you need some logical control over character removal, use this
String string = "sdsdsd";
char[] arr = string.toCharArray();
// Run loop or whatever you need
String ss = new String(arr);
If you don't need any such control, you can use what Oscar orBhesh mentioned. They are spot on.

Easiest way to remove a char from string
String str="welcome";
str=str.replaceFirst(String.valueOf(str.charAt(2)),"");//'l' will replace with ""
System.out.println(str);//output: wecome

public class RemoveCharFromString {
public static void main(String[] args) {
String output = remove("Hello", 'l');
System.out.println(output);
}
private static String remove(String input, char c) {
if (input == null || input.length() <= 1)
return input;
char[] inputArray = input.toCharArray();
char[] outputArray = new char[inputArray.length];
int outputArrayIndex = 0;
for (int i = 0; i < inputArray.length; i++) {
char p = inputArray[i];
if (p != c) {
outputArray[outputArrayIndex] = p;
outputArrayIndex++;
}
}
return new String(outputArray, 0, outputArrayIndex);
}
}

In most use-cases using StringBuilder or substring is a good approach (as already answered). However, for performance critical code, this might be a good alternative.
/**
* Delete a single character from index position 'start' from the 'target' String.
*
* ````
* deleteAt("ABC", 0) -> "BC"
* deleteAt("ABC", 1) -> "B"
* deleteAt("ABC", 2) -> "C"
* ````
*/
public static String deleteAt(final String target, final int start) {
return deleteAt(target, start, start + 1);
}
/**
* Delete the characters from index position 'start' to 'end' from the 'target' String.
*
* ````
* deleteAt("ABC", 0, 1) -> "BC"
* deleteAt("ABC", 0, 2) -> "C"
* deleteAt("ABC", 1, 3) -> "A"
* ````
*/
public static String deleteAt(final String target, final int start, int end) {
final int targetLen = target.length();
if (start < 0) {
throw new IllegalArgumentException("start=" + start);
}
if (end > targetLen || end < start) {
throw new IllegalArgumentException("end=" + end);
}
if (start == 0) {
return end == targetLen ? "" : target.substring(end);
} else if (end == targetLen) {
return target.substring(0, start);
}
final char[] buffer = new char[targetLen - end + start];
target.getChars(0, start, buffer, 0);
target.getChars(end, targetLen, buffer, start);
return new String(buffer);
}

*You can delete string value use the StringBuilder and deletecharAt.
String s1 = "aabc";
StringBuilder sb = new StringBuilder(s1);
for(int i=0;i<sb.length();i++)
{
char temp = sb.charAt(0);
if(sb.indexOf(temp+"")!=1)
{
sb.deleteCharAt(sb.indexOf(temp+""));
}
}

To Remove a Single character from The Given String please find my method hope it will be usefull. i have used str.replaceAll to remove the string but their are many ways to remove a character from a given string but i prefer replaceall method.
Code For Remove Char:
import java.util.ArrayList;
import java.util.Collection;
import java.util.Collections;
public class Removecharacter
{
public static void main(String[] args)
{
String result = removeChar("Java", 'a');
String result1 = removeChar("Edition", 'i');
System.out.println(result + " " + result1);
}
public static String removeChar(String str, char c) {
if (str == null)
{
return null;
}
else
{
return str.replaceAll(Character.toString(c), "");
}
}
}
Console image :
please find The Attached image of console,
Thanks For Asking. :)

public static String removechar(String fromString, Character character) {
int indexOf = fromString.indexOf(character);
if(indexOf==-1)
return fromString;
String front = fromString.substring(0, indexOf);
String back = fromString.substring(indexOf+1, fromString.length());
return front+back;
}

BufferedReader input=new BufferedReader(new InputStreamReader(System.in));
String line1=input.readLine();
String line2=input.readLine();
char[] a=line2.toCharArray();
char[] b=line1.toCharArray();
loop: for(int t=0;t<a.length;t++) {
char a1=a[t];
for(int t1=0;t1<b.length;t1++) {
char b1=b[t1];
if(a1==b1) {
StringBuilder sb = new StringBuilder(line1);
sb.deleteCharAt(t1);
line1=sb.toString();
b=line1.toCharArray();
list.add(a1);
continue loop;
}
}

When I have these kinds of questions I always ask: "what would the Java Gurus do?" :)
And I'd answer that, in this case, by looking at the implementation of String.trim().
Here's an extrapolation of that implementation that allows for more trim characters to be used.
However, note that original trim actually removes all chars that are <= ' ', so you may have to combine this with the original to get the desired result.
String trim(String string, String toTrim) {
// input checks removed
if (toTrim.length() == 0)
return string;
final char[] trimChars = toTrim.toCharArray();
Arrays.sort(trimChars);
int start = 0;
int end = string.length();
while (start < end &&
Arrays.binarySearch(trimChars, string.charAt(start)) >= 0)
start++;
while (start < end &&
Arrays.binarySearch(trimChars, string.charAt(end - 1)) >= 0)
end--;
return string.substring(start, end);
}

public String missingChar(String str, int n) {
String front = str.substring(0, n);
// Start this substring at n+1 to omit the char.
// Can also be shortened to just str.substring(n+1)
// which goes through the end of the string.
String back = str.substring(n+1, str.length());
return front + back;
}

I just implemented this utility class that removes a char or a group of chars from a String. I think it's fast because doesn't use Regexp. I hope that it helps someone!
package your.package.name;
/**
* Utility class that removes chars from a String.
*
*/
public class RemoveChars {
public static String remove(String string, String remove) {
return new String(remove(string.toCharArray(), remove.toCharArray()));
}
public static char[] remove(final char[] chars, char[] remove) {
int count = 0;
char[] buffer = new char[chars.length];
for (int i = 0; i < chars.length; i++) {
boolean include = true;
for (int j = 0; j < remove.length; j++) {
if ((chars[i] == remove[j])) {
include = false;
break;
}
}
if (include) {
buffer[count++] = chars[i];
}
}
char[] output = new char[count];
System.arraycopy(buffer, 0, output, 0, count);
return output;
}
/**
* For tests!
*/
public static void main(String[] args) {
String string = "THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG";
String remove = "AEIOU";
System.out.println();
System.out.println("Remove AEIOU: " + string);
System.out.println("Result: " + RemoveChars.remove(string, remove));
}
}
This is the output:
Remove AEIOU: THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG
Result: TH QCK BRWN FX JMPS VR TH LZY DG

For example if you want to calculate how many a's are there in the String, you can do it like this:
if (string.contains("a"))
{
numberOf_a++;
string = string.replaceFirst("a", "");
}

Algorithm for duplicated but overlapping strings

I need to write a method where I'm given a string s and I need to return the shortest string which contains s as a contiguous substring twice.
However two occurrences of s may overlap. For example,
aba returns ababa
xxxxx returns xxxxxx
abracadabra returns abracadabracadabra
My code so far is this:
import java.util.Scanner;
public class TwiceString {
public static String getShortest(String s) {
int index = -1, i, j = s.length() - 1;
char[] arr = s.toCharArray();
String res = s;
for (i = 0; i < j; i++, j--) {
if (arr[i] == arr[j]) {
index = i;
} else {
break;
}
}
if (index != -1) {
for (i = index + 1; i <= j; i++) {
String tmp = new String(arr, i, i);
res = res + tmp;
}
} else {
res = res + res;
}
return res;
}
public static void main(String args[]) {
Scanner inp = new Scanner(System.in);
System.out.println("Enter the string: ");
String word = inp.next();
System.out.println("The requires shortest string is " + getShortest(word));
}
}
I know I'm probably wrong at the algorithmic level rather than at the coding level. What should be my algorithm?

Use a suffix tree. In particular, after you've constructed the tree for s, go to the leaf representing the whole string and walk up until you see another end-of-string marker. This will be the leaf of the longest suffix that is also a prefix of s.

As #phs already said, part of the problem can be translated to "find the longest prefix of s that is also a suffix of s" and a solution without a tree may be this:
public static String getShortest(String s) {
int i = s.length();
while(i > 0 && !s.endsWith(s.substring(0, --i)))
;
return s + s.substring(i);
}

Once you've found your index, and even if it's -1, you just need to append to the original string the substring going from index + 1 (since index is the last matching character index) to the end of the string. There's a method in String to get this substring.

i think you should have a look at the Knuth-Morris-Pratt algorithm, the partial match table it uses is pretty much what you need (and by the way it's a very nice algorithm ;)

If your input string s is, say, "abcde" you can easily build a regex like the following (notice that the last character "e" is missing!):
a(b(c(d)?)?)?$
and run it on the string s. This will return the starting position of the trailing repeated substring. You would then just append the missing part (i.e. the last N-M characters of s, where N is the length of s and M is the length of the match), e.g.
aba
^ match "a"; append the missing "ba"
xxxxxx
^ match "xxxxx"; append the missing "x"
abracadabra
^ match "abra"; append the missing "cadabra"
nooverlap
--> no match; append "nooverlap"

From my understanding you want to do this:
input: dog
output: dogdog
--------------
input: racecar
output: racecaracecar
So this is how i would do that:
public String change(String input)
{
StringBuilder outputBuilder = new StringBuilder(input);
int patternLocation = input.length();
for(int x = 1;x < input.length();x++)
{
StringBuilder check = new StringBuilder(input);
for(int y = 0; y < x;y++)
check.deleteCharAt(check.length() - 1);
if(input.endsWith(check.toString()))
{
patternLocation = x;
break;
}
}
outputBuilder.delete(0, input.length() - patternLocation);
return outputBuilder.toString();
}
Hope this helped!