Develop Reference

A function which increments a string, to create a new string

write a function which increments a string, to create a new string.
If the string already ends with a number, the number should be incremented by 1.
If the string does not end with a number. the number 1 should be appended to the new string.
Examples:
foo - foo1
foobar23 - foobar24
foo0042 - foo0043
foo9 - foo10
foo099 - foo100
Attention: If the number has leading zeros the amount of digits should be considered.
The program passed tests on the CodeWars platform, except for one
For input string: "1712031362069931272877416673"
she falls on it
at java.base/java.lang.NumberFormatException.forInputString(NumberFormatException.java:67)
but in IJ it works correctly ...
Any idea why?
import java.math.BigInteger;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Main {
public static void main(String[] args) {
System.out.println(incrementString("foo001"));
System.out.println(incrementString("33275375531813209960"));
System.out.println(incrementString("0000004617702678077138438340108"));
}
public static String incrementString(String str) {
boolean isNumeric = str.chars().allMatch( Character::isDigit );
if(str.isEmpty())
return "1";
else if (isNumeric) {
BigInteger b = new BigInteger(str);
return String.format("%0"+str.length() + "d",b.add(BigInteger.valueOf(1)));
}
String timeRegex = "(.*)(\\D)([0-9]*)";
Pattern pattern = Pattern.compile(timeRegex);
Matcher matcher = pattern.matcher(str);
if (matcher.matches()) {
String sec = matcher.group(3);
StringBuilder sb = new StringBuilder();
if (sec.isEmpty()) {
sec = "0";
return str + sb+(Integer.parseInt(sec) + 1);
} else {
int length = String.valueOf(Integer.parseInt(sec) + 1).length();
if (sec.length() > length) {
for (int i = length; i < sec.length(); i++) {
sb.append("0");
}
}
return str.substring(0,str.length() - sec.length()) + String.format("%0"+sec.length() + "d",Integer.parseInt(sec)+1);
}
}
else
return "";
}
}

The issue is with Integer.parseInt(sec) when trying to parse a value too long to fix in a int (max is 2 billion)
You need to use BigInteger everywhere, also there is much useless code. You can also capture the zeros in the first group of the regex, so you don't have leading zeros to take care
public static String incrementString(String str) {
boolean isNumeric = str.chars().allMatch(Character::isDigit);
if (str.isEmpty()) {
return "1";
} else if (isNumeric) {
BigInteger b = new BigInteger(str);
return String.format("%0" + str.length() + "d", b.add(BigInteger.ONE));
}
String timeRegex = "(.*\\D0*)([1-9][0-9]*)";
Matcher matcher = Pattern.compile(timeRegex).matcher(str);
if (matcher.matches()) {
String sec = matcher.group(2);
if (sec.isEmpty()) {
return str + 1;
} else {
BigInteger new_value = new BigInteger(sec).add(BigInteger.ONE);
return matcher.group(1) + new_value;
}
}
return "";
}

How would you solve this problem by hand? I'll bet you wouldn't require a calculator.
The way I would do would be to just look at the last character in the string:
If the string is empty or the last character is not a digit, append the character 1.
If the last character is one of the digits 0 to 8, change it to the next digit.
If the last character is the digit 9:
Remove all the trailing 9s
Apply whichever of (1) or (2) above is appropriate.
Append the same number of 0s as the number of 9s you removed.
You can implement that simple algorithm in a few lines, without BigInteger and without regexes.
This seems to work, although I didn't test it thoroughly with different Unicode scripts (and I'm really not a Java programmer):
public static String incrementString(String str) {
if (str.isEmpty())
return "1";
char lastChar = str.charAt(str.length()-1);
if (!Character.isDigit(lastChar))
return str + "1";
String prefix = str.substring(0, str.length()-1);
if (Character.digit(lastChar, 10) != 9)
return prefix + (char)(lastChar + 1);
return incrementString(prefix) + (char)(lastChar - 9);
}

How to split a string in format AB123 --> AB 123? Java

I have a string in format AB123. I want to split it between the AB and 123 so AB123 becomes AB 123. The contents of the string can differ but the format stays the same. Is there a way to do this?

Following up with the latest information you provided (2 letters then 3 numbers):
myString.subString(0, 2) + " " + myString.subString(2)
What this does: you split your input string myString at the 2nd character and append a space at this position.

Explanation: \D represents non-digit and \d represents a digit in a regular expression and I used ternary operation in the regex to split charter to the number.
String string = "AB123";
String[] split = string.split("(?<=\\D)(?=\\d)");
System.out.println(split[0]+" "+split[1]);

Try
String a = "abcd1234";
int i;
for(i = 0; i < a.length(); i++){
char c = a.charAt(i);
if( '0' <= c && c <= '9' )
break;
}
String alphaPart = a.substring(0, i);
String numberPart = a.substring(i);
Hope this helps

Although I would personally use the method provided in #RakeshMothukur's answer, since it also works when the letter or digit counts increase/decrease later on, I wanted to provide an additional method to insert the space between the two letters and three digits:
String str = "AB123";
StringBuilder sb = new StringBuilder(str);
sb.insert(2, " "); // Insert a space at 0-based index 2; a.k.a. after the first 2 characters
String result = sb.toString(); // Convert the StringBuilder back to a String
Try it online.

Here you go. I wrote it in very simple way to make things clear.
What it does is : After it takes user input, it converts the string into Char array and it checks single character if its INT or non INT.
In each iteration it compares the data type with the prev character and prints accordingly.
Alternate Solutions
1) Using ASCII range (difficulty = easy)
2) Override a method and check 2 variables at a time. (difficulty = Intermediate)
import org.omg.CORBA.INTERNAL;
import java.io.InputStreamReader;
import java.util.*;
import java.io.BufferedReader;
public class Main {
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
char[] s = br.readLine().toCharArray();
int prevflag, flag = 0;
for (int i = 0; i < s.length; i++) {
int a = Character.getNumericValue(s[i]);
String b = String.valueOf(s[i]);
prevflag = flag;
flag = checktype(a, b);
if ((prevflag == flag) || (i == 0))
System.out.print(s[i]);
else
System.out.print(" " + s[i]);
}
}
public static int checktype(int x, String y) {
int flag = 0;
if (String.valueOf(x).equals(y))
flag = 1; // INT
else
flag = 2; // non INT
return flag;
}
}

I was waiting for a compile to finish before heading out, so threw together a slightly over-engineered example with basic error checking and a test.
import java.text.ParseException;
import java.util.LinkedList;
public class Main {
static public class ParsedData {
public final String prefix;
public final Integer number;
public ParsedData(String _prefix, Integer _number) {
prefix = _prefix;
number = _number;
}
#Override
public String toString() {
return prefix + "\t" + number.toString();
}
}
static final String TEST_DATA[] = {"AB123", "JX7272", "FX402", "ADF123", "JD3Q2", "QB778"};
public static void main(String[] args) {
parseDataArray(TEST_DATA);
}
public static ParsedData[] parseDataArray(String[] inputs) {
LinkedList<ParsedData> results = new LinkedList<ParsedData>();
for (String s : TEST_DATA) {
try {
System.out.println("Parsing: " + s);
if (s.length() != 5) throw new ParseException("Input Length incorrect: " + s.length(), 0);
String _prefix = s.substring(0, 2);
Integer _num = Integer.parseInt(s.substring(2));
results.add(new ParsedData(_prefix, _num));
} catch (ParseException | NumberFormatException e) {
System.out.printf("\"%s\", %s\n", s, e.toString());
}
}
return results.toArray(new ParsedData[results.size()]);
}
}

Reverse string printing method

I am trying to solve the following problem but how do write the method that accepts String as an argument?
Write a method named printReverse that accepts a String as an
argument and prints the characters in the opposite order. If the empty
string is passed as an argument, the method should produce no output.
Be sure to write a main method that convincingly demonstrates your
program in action. Do not use the reverse method of the
StringBuilder or StringBuffer class!
So far I have solved it in a easier manner:
import java.util.Scanner;
class ReverseString {
public static void main(String args[]) {
String original, reverse = "";
Scanner in = new Scanner(System.in);
System.out.println("Enter a string to reverse");
original = in.nextLine();
int length = original.length();
for (int i = length - 1; i >= 0; i--)
reverse = reverse + original.charAt(i);
System.out.println("Reverse of entered string is: " + reverse);
}
}

I highly recommend you to go through a basic tutorial.
You can simply do:
private static String myReverse(String str) {
String reverse = "";
int length = str.length();
for( int i = length - 1 ; i >= 0 ; i-- ) {
reverse = reverse + str.charAt(i);
}
return reverse;
}
And in your main, you simply:
String reversed = myReverse(in.nextLine());
Note that the method is static because you're referring to it from a static manner (main method). If you don't want it to be static, you'll have to access it via an object.
Also note that it's a good practice to always have curly brackets for for loops, even if it contains a single line.

how do write the method that accepts String as an argument?
public static String reverse(String forward) {
char[] strChar = forward.toCharArray();
String reverse = "";
for( int i = strChar.length - 1 ; i >= 0 ; i-- )
reverse = reverse + strChar[i];
return reverse;
}
But for large string appending character with + operator can be inefficient. And reversing string with above approach will result in wrong for uni-code mismatches. As it reverse the code units but not character. There is actually a built-in support available to reverse a string using StringBuilder which works correctly:
public static String reverse(String forward) {
StringBuilder builder = new StringBuilder(forward);
String reverse = builder.reverse().toString();
return reverse;
}

Something like this:
public class StringUtils {
public static String reverse(String forward) {
String result = "";
// Put your code here
return result;
}
}

Using Java 9 you can implement something like this. This code works with both regular characters and surrogate pairs:
public static void printReverse(String str) {
// character code points
str.codePoints()
// character as string
.mapToObj(Character::toString)
// concatenate in reverse order
.reduce((a, b) -> b + a)
// output
.ifPresent(System.out::println);
}
public static void main(String[] args) {
// regular characters
printReverse("lorem ipsum");
// surrogate pairs
printReverse("\uD835\uDD43\uD835\uDD46R\uD835\uDD3C\uD835\uDD44" +
" \uD835\uDD40P\uD835\uDD4A\uD835\uDD4C\uD835\uDD44");
}
Output:
muspi merol
𝕄𝕌𝕊P𝕀 𝕄𝔼R𝕆𝕃
See also: Is there any other way to remove all whitespaces in a string?

Try this:
private static String reverseString(String str) {
String revString = "";
for (int i = str.length() - 1; i >= 0; i--) {
revString = revString + str.charAt(i);
}
return revString;
}

package dowhile;
public class Dowhile {
public static void main(String[] args) {
// TODO code application logic here
String message = "i love java programming";
int msglength = message.length();
int index = msglength - 1;
while (index >= 0) {
System.out.print(message.charAt(index));
index--;
}
}
}
Output:
gnimmargorp avaj evol i

private static void printReverse(String org) {
StringBuffer buffer = new StringBuffer(org);
String reversedStr = buffer.reverse().toString();
System.out.println("The reverse of the string \""
+ str + "\" is \"" + reversedStr + "\".");
}
in the main call the function
printReverse(original);

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", "");
}

How to capitalize the first character of each word in a string

Is there a function built into Java that capitalizes the first character of each word in a String, and does not affect the others?
Examples:
jon skeet -> Jon Skeet
miles o'Brien -> Miles O'Brien (B remains capital, this rules out Title Case)
old mcdonald -> Old Mcdonald*
*(Old McDonald would be find too, but I don't expect it to be THAT smart.)
A quick look at the Java String Documentation reveals only toUpperCase() and toLowerCase(), which of course do not provide the desired behavior. Naturally, Google results are dominated by those two functions. It seems like a wheel that must have been invented already, so it couldn't hurt to ask so I can use it in the future.

WordUtils.capitalize(str) (from apache commons-text)
(Note: if you need "fOO BAr" to become "Foo Bar", then use capitalizeFully(..) instead)

If you're only worried about the first letter of the first word being capitalized:
private String capitalize(final String line) {
return Character.toUpperCase(line.charAt(0)) + line.substring(1);
}

The following method converts all the letters into upper/lower case, depending on their position near a space or other special chars.
public static String capitalizeString(String string) {
char[] chars = string.toLowerCase().toCharArray();
boolean found = false;
for (int i = 0; i < chars.length; i++) {
if (!found && Character.isLetter(chars[i])) {
chars[i] = Character.toUpperCase(chars[i]);
found = true;
} else if (Character.isWhitespace(chars[i]) || chars[i]=='.' || chars[i]=='\'') { // You can add other chars here
found = false;
}
}
return String.valueOf(chars);
}

Try this very simple way
example givenString="ram is good boy"
public static String toTitleCase(String givenString) {
String[] arr = givenString.split(" ");
StringBuffer sb = new StringBuffer();
for (int i = 0; i < arr.length; i++) {
sb.append(Character.toUpperCase(arr[i].charAt(0)))
.append(arr[i].substring(1)).append(" ");
}
return sb.toString().trim();
}
Output will be: Ram Is Good Boy

I made a solution in Java 8 that is IMHO more readable.
public String firstLetterCapitalWithSingleSpace(final String words) {
return Stream.of(words.trim().split("\\s"))
.filter(word -> word.length() > 0)
.map(word -> word.substring(0, 1).toUpperCase() + word.substring(1))
.collect(Collectors.joining(" "));
}
The Gist for this solution can be found here: https://gist.github.com/Hylke1982/166a792313c5e2df9d31

String toBeCapped = "i want this sentence capitalized";
String[] tokens = toBeCapped.split("\\s");
toBeCapped = "";
for(int i = 0; i < tokens.length; i++){
char capLetter = Character.toUpperCase(tokens[i].charAt(0));
toBeCapped += " " + capLetter + tokens[i].substring(1);
}
toBeCapped = toBeCapped.trim();

I've written a small Class to capitalize all the words in a String.
Optional multiple delimiters, each one with its behavior (capitalize before, after, or both, to handle cases like O'Brian);
Optional Locale;
Don't breaks with Surrogate Pairs.
LIVE DEMO
Output:
====================================
SIMPLE USAGE
====================================
Source: cApItAlIzE this string after WHITE SPACES
Output: Capitalize This String After White Spaces
====================================
SINGLE CUSTOM-DELIMITER USAGE
====================================
Source: capitalize this string ONLY before'and''after'''APEX
Output: Capitalize this string only beforE'AnD''AfteR'''Apex
====================================
MULTIPLE CUSTOM-DELIMITER USAGE
====================================
Source: capitalize this string AFTER SPACES, BEFORE'APEX, and #AFTER AND BEFORE# NUMBER SIGN (#)
Output: Capitalize This String After Spaces, BeforE'apex, And #After And BeforE# Number Sign (#)
====================================
SIMPLE USAGE WITH CUSTOM LOCALE
====================================
Source: Uniforming the first and last vowels (different kind of 'i's) of the Turkish word D[İ]YARBAK[I]R (DİYARBAKIR)
Output: Uniforming The First And Last Vowels (different Kind Of 'i's) Of The Turkish Word D[i]yarbak[i]r (diyarbakir)
====================================
SIMPLE USAGE WITH A SURROGATE PAIR
====================================
Source: ab 𐐂c de à
Output: Ab 𐐪c De À
Note: first letter will always be capitalized (edit the source if you don't want that).
Please share your comments and help me to found bugs or to improve the code...
Code:
import java.util.ArrayList;
import java.util.Date;
import java.util.List;
import java.util.Locale;
public class WordsCapitalizer {
public static String capitalizeEveryWord(String source) {
return capitalizeEveryWord(source,null,null);
}
public static String capitalizeEveryWord(String source, Locale locale) {
return capitalizeEveryWord(source,null,locale);
}
public static String capitalizeEveryWord(String source, List<Delimiter> delimiters, Locale locale) {
char[] chars;
if (delimiters == null || delimiters.size() == 0)
delimiters = getDefaultDelimiters();
// If Locale specified, i18n toLowerCase is executed, to handle specific behaviors (eg. Turkish dotted and dotless 'i')
if (locale!=null)
chars = source.toLowerCase(locale).toCharArray();
else
chars = source.toLowerCase().toCharArray();
// First charachter ALWAYS capitalized, if it is a Letter.
if (chars.length>0 && Character.isLetter(chars[0]) && !isSurrogate(chars[0])){
chars[0] = Character.toUpperCase(chars[0]);
}
for (int i = 0; i < chars.length; i++) {
if (!isSurrogate(chars[i]) && !Character.isLetter(chars[i])) {
// Current char is not a Letter; gonna check if it is a delimitrer.
for (Delimiter delimiter : delimiters){
if (delimiter.getDelimiter()==chars[i]){
// Delimiter found, applying rules...
if (delimiter.capitalizeBefore() && i>0
&& Character.isLetter(chars[i-1]) && !isSurrogate(chars[i-1]))
{ // previous character is a Letter and I have to capitalize it
chars[i-1] = Character.toUpperCase(chars[i-1]);
}
if (delimiter.capitalizeAfter() && i<chars.length-1
&& Character.isLetter(chars[i+1]) && !isSurrogate(chars[i+1]))
{ // next character is a Letter and I have to capitalize it
chars[i+1] = Character.toUpperCase(chars[i+1]);
}
break;
}
}
}
}
return String.valueOf(chars);
}
private static boolean isSurrogate(char chr){
// Check if the current character is part of an UTF-16 Surrogate Pair.
// Note: not validating the pair, just used to bypass (any found part of) it.
return (Character.isHighSurrogate(chr) || Character.isLowSurrogate(chr));
}
private static List<Delimiter> getDefaultDelimiters(){
// If no delimiter specified, "Capitalize after space" rule is set by default.
List<Delimiter> delimiters = new ArrayList<Delimiter>();
delimiters.add(new Delimiter(Behavior.CAPITALIZE_AFTER_MARKER, ' '));
return delimiters;
}
public static class Delimiter {
private Behavior behavior;
private char delimiter;
public Delimiter(Behavior behavior, char delimiter) {
super();
this.behavior = behavior;
this.delimiter = delimiter;
}
public boolean capitalizeBefore(){
return (behavior.equals(Behavior.CAPITALIZE_BEFORE_MARKER)
|| behavior.equals(Behavior.CAPITALIZE_BEFORE_AND_AFTER_MARKER));
}
public boolean capitalizeAfter(){
return (behavior.equals(Behavior.CAPITALIZE_AFTER_MARKER)
|| behavior.equals(Behavior.CAPITALIZE_BEFORE_AND_AFTER_MARKER));
}
public char getDelimiter() {
return delimiter;
}
}
public static enum Behavior {
CAPITALIZE_AFTER_MARKER(0),
CAPITALIZE_BEFORE_MARKER(1),
CAPITALIZE_BEFORE_AND_AFTER_MARKER(2);
private int value;
private Behavior(int value) {
this.value = value;
}
public int getValue() {
return value;
}
}

Using org.apache.commons.lang.StringUtils makes it very simple.
capitalizeStr = StringUtils.capitalize(str);

From Java 9+
you can use String::replaceAll like this :
public static void upperCaseAllFirstCharacter(String text) {
String regex = "\\b(.)(.*?)\\b";
String result = Pattern.compile(regex).matcher(text).replaceAll(
matche -> matche.group(1).toUpperCase() + matche.group(2)
);
System.out.println(result);
}
Example :
upperCaseAllFirstCharacter("hello this is Just a test");
Outputs
Hello This Is Just A Test

With this simple code:
String example="hello";
example=example.substring(0,1).toUpperCase()+example.substring(1, example.length());
System.out.println(example);
Result: Hello

I'm using the following function. I think it is faster in performance.
public static String capitalize(String text){
String c = (text != null)? text.trim() : "";
String[] words = c.split(" ");
String result = "";
for(String w : words){
result += (w.length() > 1? w.substring(0, 1).toUpperCase(Locale.US) + w.substring(1, w.length()).toLowerCase(Locale.US) : w) + " ";
}
return result.trim();
}

Use the Split method to split your string into words, then use the built in string functions to capitalize each word, then append together.
Pseudo-code (ish)
string = "the sentence you want to apply caps to";
words = string.split(" ")
string = ""
for(String w: words)
//This line is an easy way to capitalize a word
word = word.toUpperCase().replace(word.substring(1), word.substring(1).toLowerCase())
string += word
In the end string looks something like
"The Sentence You Want To Apply Caps To"

This might be useful if you need to capitalize titles. It capitalizes each substring delimited by " ", except for specified strings such as "a" or "the". I haven't ran it yet because it's late, should be fine though. Uses Apache Commons StringUtils.join() at one point. You can substitute it with a simple loop if you wish.
private static String capitalize(String string) {
if (string == null) return null;
String[] wordArray = string.split(" "); // Split string to analyze word by word.
int i = 0;
lowercase:
for (String word : wordArray) {
if (word != wordArray[0]) { // First word always in capital
String [] lowercaseWords = {"a", "an", "as", "and", "although", "at", "because", "but", "by", "for", "in", "nor", "of", "on", "or", "so", "the", "to", "up", "yet"};
for (String word2 : lowercaseWords) {
if (word.equals(word2)) {
wordArray[i] = word;
i++;
continue lowercase;
}
}
}
char[] characterArray = word.toCharArray();
characterArray[0] = Character.toTitleCase(characterArray[0]);
wordArray[i] = new String(characterArray);
i++;
}
return StringUtils.join(wordArray, " "); // Re-join string
}

public static String toTitleCase(String word){
return Character.toUpperCase(word.charAt(0)) + word.substring(1);
}
public static void main(String[] args){
String phrase = "this is to be title cased";
String[] splitPhrase = phrase.split(" ");
String result = "";
for(String word: splitPhrase){
result += toTitleCase(word) + " ";
}
System.out.println(result.trim());
}

1. Java 8 Streams
public static String capitalizeAll(String str) {
if (str == null || str.isEmpty()) {
return str;
}
return Arrays.stream(str.split("\\s+"))
.map(t -> t.substring(0, 1).toUpperCase() + t.substring(1))
.collect(Collectors.joining(" "));
}
Examples:
System.out.println(capitalizeAll("jon skeet")); // Jon Skeet
System.out.println(capitalizeAll("miles o'Brien")); // Miles O'Brien
System.out.println(capitalizeAll("old mcdonald")); // Old Mcdonald
System.out.println(capitalizeAll(null)); // null
For foo bAR to Foo Bar, replace the map() method with the following:
.map(t -> t.substring(0, 1).toUpperCase() + t.substring(1).toLowerCase())
2. String.replaceAll() (Java 9+)
ublic static String capitalizeAll(String str) {
if (str == null || str.isEmpty()) {
return str;
}
return Pattern.compile("\\b(.)(.*?)\\b")
.matcher(str)
.replaceAll(match -> match.group(1).toUpperCase() + match.group(2));
}
Examples:
System.out.println(capitalizeAll("12 ways to learn java")); // 12 Ways To Learn Java
System.out.println(capitalizeAll("i am atta")); // I Am Atta
System.out.println(capitalizeAll(null)); // null
3. Apache Commons Text
System.out.println(WordUtils.capitalize("love is everywhere")); // Love Is Everywhere
System.out.println(WordUtils.capitalize("sky, sky, blue sky!")); // Sky, Sky, Blue Sky!
System.out.println(WordUtils.capitalize(null)); // null
For titlecase:
System.out.println(WordUtils.capitalizeFully("fOO bAR")); // Foo Bar
System.out.println(WordUtils.capitalizeFully("sKy is BLUE!")); // Sky Is Blue!
For details, checkout this tutorial.

BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.println("Enter the sentence : ");
try
{
String str = br.readLine();
char[] str1 = new char[str.length()];
for(int i=0; i<str.length(); i++)
{
str1[i] = Character.toLowerCase(str.charAt(i));
}
str1[0] = Character.toUpperCase(str1[0]);
for(int i=0;i<str.length();i++)
{
if(str1[i] == ' ')
{
str1[i+1] = Character.toUpperCase(str1[i+1]);
}
System.out.print(str1[i]);
}
}
catch(Exception e)
{
System.err.println("Error: " + e.getMessage());
}

I decided to add one more solution for capitalizing words in a string:
words are defined here as adjacent letter-or-digit characters;
surrogate pairs are provided as well;
the code has been optimized for performance; and
it is still compact.
Function:
public static String capitalize(String string) {
final int sl = string.length();
final StringBuilder sb = new StringBuilder(sl);
boolean lod = false;
for(int s = 0; s < sl; s++) {
final int cp = string.codePointAt(s);
sb.appendCodePoint(lod ? Character.toLowerCase(cp) : Character.toUpperCase(cp));
lod = Character.isLetterOrDigit(cp);
if(!Character.isBmpCodePoint(cp)) s++;
}
return sb.toString();
}
Example call:
System.out.println(capitalize("An à la carte StRiNg. Surrogate pairs: 𐐪𐐪."));
Result:
An À La Carte String. Surrogate Pairs: 𐐂𐐪.

Use:
String text = "jon skeet, miles o'brien, old mcdonald";
Pattern pattern = Pattern.compile("\\b([a-z])([\\w]*)");
Matcher matcher = pattern.matcher(text);
StringBuffer buffer = new StringBuffer();
while (matcher.find()) {
matcher.appendReplacement(buffer, matcher.group(1).toUpperCase() + matcher.group(2));
}
String capitalized = matcher.appendTail(buffer).toString();
System.out.println(capitalized);

There are many way to convert the first letter of the first word being capitalized. I have an idea. It's very simple:
public String capitalize(String str){
/* The first thing we do is remove whitespace from string */
String c = str.replaceAll("\\s+", " ");
String s = c.trim();
String l = "";
for(int i = 0; i < s.length(); i++){
if(i == 0){ /* Uppercase the first letter in strings */
l += s.toUpperCase().charAt(i);
i++; /* To i = i + 1 because we don't need to add
value i = 0 into string l */
}
l += s.charAt(i);
if(s.charAt(i) == 32){ /* If we meet whitespace (32 in ASCII Code is whitespace) */
l += s.toUpperCase().charAt(i+1); /* Uppercase the letter after whitespace */
i++; /* Yo i = i + 1 because we don't need to add
value whitespace into string l */
}
}
return l;
}

package com.test;
/**
* #author Prasanth Pillai
* #date 01-Feb-2012
* #description : Below is the test class details
*
* inputs a String from a user. Expect the String to contain spaces and alphanumeric characters only.
* capitalizes all first letters of the words in the given String.
* preserves all other characters (including spaces) in the String.
* displays the result to the user.
*
* Approach : I have followed a simple approach. However there are many string utilities available
* for the same purpose. Example : WordUtils.capitalize(str) (from apache commons-lang)
*
*/
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Test {
public static void main(String[] args) throws IOException{
System.out.println("Input String :\n");
InputStreamReader converter = new InputStreamReader(System.in);
BufferedReader in = new BufferedReader(converter);
String inputString = in.readLine();
int length = inputString.length();
StringBuffer newStr = new StringBuffer(0);
int i = 0;
int k = 0;
/* This is a simple approach
* step 1: scan through the input string
* step 2: capitalize the first letter of each word in string
* The integer k, is used as a value to determine whether the
* letter is the first letter in each word in the string.
*/
while( i < length){
if (Character.isLetter(inputString.charAt(i))){
if ( k == 0){
newStr = newStr.append(Character.toUpperCase(inputString.charAt(i)));
k = 2;
}//this else loop is to avoid repeatation of the first letter in output string
else {
newStr = newStr.append(inputString.charAt(i));
}
} // for the letters which are not first letter, simply append to the output string.
else {
newStr = newStr.append(inputString.charAt(i));
k=0;
}
i+=1;
}
System.out.println("new String ->"+newStr);
}
}

Here is a simple function
public static String capEachWord(String source){
String result = "";
String[] splitString = source.split(" ");
for(String target : splitString){
result += Character.toUpperCase(target.charAt(0))
+ target.substring(1) + " ";
}
return result.trim();
}

This is just another way of doing it:
private String capitalize(String line)
{
StringTokenizer token =new StringTokenizer(line);
String CapLine="";
while(token.hasMoreTokens())
{
String tok = token.nextToken().toString();
CapLine += Character.toUpperCase(tok.charAt(0))+ tok.substring(1)+" ";
}
return CapLine.substring(0,CapLine.length()-1);
}

Reusable method for intiCap:
public class YarlagaddaSireeshTest{
public static void main(String[] args) {
String FinalStringIs = "";
String testNames = "sireesh yarlagadda test";
String[] name = testNames.split("\\s");
for(String nameIs :name){
FinalStringIs += getIntiCapString(nameIs) + ",";
}
System.out.println("Final Result "+ FinalStringIs);
}
public static String getIntiCapString(String param) {
if(param != null && param.length()>0){
char[] charArray = param.toCharArray();
charArray[0] = Character.toUpperCase(charArray[0]);
return new String(charArray);
}
else {
return "";
}
}
}

Here is my solution.
I ran across this problem tonight and decided to search it. I found an answer by Neelam Singh that was almost there, so I decided to fix the issue (broke on empty strings) and caused a system crash.
The method you are looking for is named capString(String s) below.
It turns "It's only 5am here" into "It's Only 5am Here".
The code is pretty well commented, so enjoy.
package com.lincolnwdaniel.interactivestory.model;
public class StringS {
/**
* #param s is a string of any length, ideally only one word
* #return a capitalized string.
* only the first letter of the string is made to uppercase
*/
public static String capSingleWord(String s) {
if(s.isEmpty() || s.length()<2) {
return Character.toUpperCase(s.charAt(0))+"";
}
else {
return Character.toUpperCase(s.charAt(0)) + s.substring(1);
}
}
/**
*
* #param s is a string of any length
* #return a title cased string.
* All first letter of each word is made to uppercase
*/
public static String capString(String s) {
// Check if the string is empty, if it is, return it immediately
if(s.isEmpty()){
return s;
}
// Split string on space and create array of words
String[] arr = s.split(" ");
// Create a string buffer to hold the new capitalized string
StringBuffer sb = new StringBuffer();
// Check if the array is empty (would be caused by the passage of s as an empty string [i.g "" or " "],
// If it is, return the original string immediately
if( arr.length < 1 ){
return s;
}
for (int i = 0; i < arr.length; i++) {
sb.append(Character.toUpperCase(arr[i].charAt(0)))
.append(arr[i].substring(1)).append(" ");
}
return sb.toString().trim();
}
}

Here we go for perfect first char capitalization of word
public static void main(String[] args) {
String input ="my name is ranjan";
String[] inputArr = input.split(" ");
for(String word : inputArr) {
System.out.println(word.substring(0, 1).toUpperCase()+word.substring(1,word.length()));
}
}
}
//Output : My Name Is Ranjan

For those of you using Velocity in your MVC, you can use the capitalizeFirstLetter() method from the StringUtils class.

String s="hi dude i want apple";
s = s.replaceAll("\\s+"," ");
String[] split = s.split(" ");
s="";
for (int i = 0; i < split.length; i++) {
split[i]=Character.toUpperCase(split[i].charAt(0))+split[i].substring(1);
s+=split[i]+" ";
System.out.println(split[i]);
}
System.out.println(s);

package corejava.string.intern;
import java.io.DataInputStream;
import java.util.ArrayList;
/*
* wap to accept only 3 sentences and convert first character of each word into upper case
*/
public class Accept3Lines_FirstCharUppercase {
static String line;
static String words[];
static ArrayList<String> list=new ArrayList<String>();
/**
* #param args
*/
public static void main(String[] args) throws java.lang.Exception{
DataInputStream read=new DataInputStream(System.in);
System.out.println("Enter only three sentences");
int i=0;
while((line=read.readLine())!=null){
method(line); //main logic of the code
if((i++)==2){
break;
}
}
display();
System.out.println("\n End of the program");
}
/*
* this will display all the elements in an array
*/
public static void display(){
for(String display:list){
System.out.println(display);
}
}
/*
* this divide the line of string into words
* and first char of the each word is converted to upper case
* and to an array list
*/
public static void method(String lineParam){
words=line.split("\\s");
for(String s:words){
String result=s.substring(0,1).toUpperCase()+s.substring(1);
list.add(result);
}
}
}

If you prefer Guava...
String myString = ...;
String capWords = Joiner.on(' ').join(Iterables.transform(Splitter.on(' ').omitEmptyStrings().split(myString), new Function<String, String>() {
public String apply(String input) {
return Character.toUpperCase(input.charAt(0)) + input.substring(1);
}
}));

String toUpperCaseFirstLetterOnly(String str) {
String[] words = str.split(" ");
StringBuilder ret = new StringBuilder();
for(int i = 0; i < words.length; i++) {
ret.append(Character.toUpperCase(words[i].charAt(0)));
ret.append(words[i].substring(1));
if(i < words.length - 1) {
ret.append(' ');
}
}
return ret.toString();
}