Develop Reference

How do I remove the last comma from the output in series using loop [duplicate]

So I made this to print primes between two numbers of my choice; however, it prints out a comma after the last number and I don't know how to take it off.
Example
in: 0 10
out: 2, 3, 5, 7,
I want 2,3,5,7
Scanner s = new Scanner(System.in);
int a = s.nextInt();
int b = s.nextInt();
for (int i = a; i <= b; i++){
int j;
for (j = 2; j<i; j++){
int p = i%j;
if(p==0){break;}
}
if(i == j){System.out.printf("%d,", i);}
}
}

Use a boolean to keep track of whether you've printed anything yet. Then your format string could be something like
anythingPrinted ? ",%d" : "%d"
That is, only include the comma in the format string if there's something printed.

Use a StringBuilder and write to the console at the end of your program.
StringBuilder sb = new StringBuilder();
for (int i = a; i <= b; i++){
int j;
for (j = 2; j<i; j++){
int p = i%j;
if(p==0){break;}
}
if(i == j){
// If the length of the StringBuilder is 0, no need for a comma
if(sb.length() != 0) {
sb.append(",");
}
sb.append(i);
}
}
System.out.println(sb);

This might seem like overkill, and for many cases it might be, but I have been writing a source code transcoder and I find this situation coming up a lot. Where I need commas in between values, or a prefix value which is only printed once. So I found it handy to create a class which simplifies things.
Again, you wouldn't probably want to use this if you code had one or two print loops in it, but maybe if you had more than a few. Perhaps you would remove in "on first" part if you were never going to use it.
public class FirstPrintOptions {
private PrintStream printStream;
private String onFirst;
private String remaining;
private boolean trip = false;
public FirstPrintOptions(PrintStream printStream, String onFirst, String remaining) {
this.printStream = printStream;
this.onFirst = onFirst;
this.remaining = remaining;
}
public void print() {
if (!trip) {
if (onFirst != null) {
printStream.print(onFirst);
}
trip = true;
} else {
if (remaining != null) {
printStream.print(remaining);
}
}
}
}
Then use it like this..
FirstPrintOptions firstPrintOptions = new FirstPrintOptions(System.out, null, ",");
for (int x=0;x<10;x++) {
firstPrintOptions.print();
System.out.print(x);
}
The results are..
0,1,2,3,4,5,6,7,8,9

I was testing and I came up with this. I was using compilejava.net so scanner doesn't work. I bypassed that part and just set a and b manually. Basically, it builds a string with the numbers and ends in a comma. Then it prints a substring including everything except the last comma.
import java.util.*;
public class HelloWorld {
public static void main(String[] args) {
//Scanner s = new Scanner(System.in);
int a = 2;
int b = 18;
String c = "Output = ";
for (int i = a; i <= b; i++){
int j;
for (j = 2; j<i; j++){
int p = i%j;
if(p==0){break;}
}
if(i == j){c=c+ Integer.toString(i) + ",";}
}
System.out.print(c.subSequence(0, c.length()-1));
}
}

this program for finding factors of a number
for(i=1;i<=number;i++)
{
if(number%i==0)
{
system.out.print(i);
if(i!=0)
{system.out.print(",");}
}
}
so i get the output for 10 as
1,2,5,10

Taking a long string of numbers and removing any zeroes in front of it, then returning that altered string (in Java)

So I have a public static method 'getWithoutLeadingZeroes', which gets passed a String and simply needs to return it without any zeroes prefixing the string of numbers.
Now, I know that I need to iterate through the string until I find the first non-zero char in the string, but I'm not exactly sure how to take the point where the method finds the non-zero char and start copying the remainder of the String into a new String, then returning it.
Here's what I have so far:
public static String getWithoutLeadingZeroes(String s) {
boolean notZero = false;
char[] t = new char[x];
for(int i = 0; i<s.length(); i++){
if(s.charAt(i) == 0){
notZero = false;
} else {
notZero = true;
}
if(notZero = true){
for(int j = index.charAt(i))
}
return ""; //to be completed
}
I created a boolean variable to stop the loop once it hits the non-zero char and I'm pretty positive the first half of the code is accurate, but its the creating of the new String to be returned that I'm a bit stuck on. Any suggestions would be welcome.

You are NOT stopping your loop.
You could do that using the break keyword.
And: your comparison is wrong, it should read
if (... == '0'

You could use String#substring(int beginIndex) instead:
public static String getWithoutLeadingZeroes(String s) {
int i = 0;
while (i < s.length() && s.charAt(i) == '0') i++;
return s.substring(i);
}

How about the following way?
public static String getWithoutLeadingZeroes(String s) {
while(s.startsWith('0'))
s = s.substring(1);
return s;
}

This should do it
String nmbrStr = "1001234";
String cleanedStr = nmbrStr; // this way the whole number will be
// returned if in has no leading zeroes
for (int i = 0; i < nmbrStr.length(); i++) {
if (nmbrStr.charAt(i) != '0') {
cleanedStr = nmbrStr.substring(i);
break;
}
}
System.out.println(cleanedStr);
You can also use Integer.parseInt(nmbrStr) which is a lot cleaner

A simple solution.
public static String getWithoutLeadingZeroes(String stringOfNumbers) {
return String.valueOf(Long.parseLong(stringOfNumbers));
}

Another alternative solution:
public String getWithoutLeadingZeroes(String str) {
int from = 0;
for (int i = 0; i < str.length(); i++) {
if(str.charAt(i) == '0'){
from = i;
}else{
break;
}
}
return str.substring(from+1);
}
You can then make it a little bit more robust by implementing try/catch blocks and possibly along with cases that deal with unexpected input e.g null parameter or a zero-length string etc.

Overlay Strings In Java While Skipping Spaces

I want to overlay two strings together in Java to make one string. I already have a method to do something like this, but not exactly what I want.
For example, if I tell it to overlay "Hello World" with " Hi", I will get " Hiorld".
I want to be able to overlay "Hello World" with " Hi", and get "HelloHiorld"!
Here is the method I currently am using:
public static String overlayString(String str, String overlay, int start, int end) {
if (str == null) {
return null;
}
if (overlay == null) {
overlay = "";
}
int len = str.length();
if (start < 0) {
start = 0;
}
if (start > len) {
start = len;
}
if (end < 0) {
end = 0;
}
if (end > len) {
end = len;
}
if (start > end) {
int temp = start;
start = end;
end = temp;
}
return new StringBuffer(len + start - end + overlay.length() + 1)
.append(str.substring(0, start))
.append(overlay)
.append(str.substring(end))
.toString();
}
How would this method look if I wanted the method to avoid replacing characters with spaces, but still keep the positioning of the text?

You could put the first String into a StringBuffer, use a loop to check which characters in the second String are not spaces, then use the StringBuffer's .replace() method to replace these characters.
For example,
StringBuffer sb = new StringBuffer("Hello World");
sb.replace(5, 7, "Hi");
gives "HelloHiorld"

I can thing of two ways you might be able to achieve something like this...
First...
public static String overlay1(String value, String with) {
String result = null;
if (value.length() != with.length()) {
throw new IllegalArgumentException("The two String's must be the same length");
} else {
StringBuilder sb = new StringBuilder(value);
for (int index = 0; index < value.length(); index++) {
char c = with.charAt(index);
if (!Character.isWhitespace(c)) {
sb.setCharAt(index, c);
}
}
result = sb.toString();
}
return result;
}
Which allows you to pass to Strings of the same length and will replace all the "non-whitespace" content of the second in the first, something like...
overlay1("Hello World",
" Hi "));
Or you could specify the location you want the String to be overlaied, for example...
public static String overlay2(String value, String with, int at) {
String result = null;
if (at >= value.length()) {
throw new IllegalArgumentException("Insert point is beyond length of original String");
} else {
StringBuilder sb = new StringBuilder(value);
// Assuming a straight replacement with out needing to
// check for white spaces...otherwise you can use
// the same type of loop from the first example
sb.replace(at, with.length(), with);
result = sb.toString();
}
return result;
}
overlay2("Hello World",
"Hi",
5));

Substring between two same or different delimiters (when delimiters occur multiple times)

I need to fetch a sub string that lies between two same or different delimiters. The delimiters will be occurring multiple times in the string, so i need to extract the sub-string that lies between mth occurrence of delimiter1 and nth occurrence of delimiter2.
For eg:
myString : Ron_CR7_MU^RM^_SAF_34^
What should i do here if i need to extract the sub-string that lies between 3rd occurrence of '_' and 3rd occurence of '^'?
Substring = SAF_34
Or i could look for a substring that lies between 2nd '^' and 4th '_', i.e :
Substring = _SAF
An SQL equivalent would be :
substr(myString, instr(myString, '',1,3)+1,instr(myString, '^',1,3)-1-instr(myString, '',1,3))

I would use,
public static int findNth(String text, String toFind, int count) {
int pos = -1;
do {
pos = text.indexOf(toFind, pos+1);
} while(--count > 0 && pos >= 0);
return pos;
}
int from = findNth(text, "_", 3);
int to = findNth(text, "^", 3);
String found = text.substring(from+1, to);

If you can use a solution without regex you can find indexes in your string where your resulting string needs to start and where it needs to end. Then just simply perform: myString.substring(start,end) to get your result.
Biggest problem is to find start and end. To do it you can repeat this N (M) times:
int pos = indexOf(delimiterX)
myString = myString.substring(pos) //you may want to work on copy of myString
Hope you get an idea.

You could create a little method that simply hunts for such substrings between delimiters sequentially, using (as noted) String.indexOf(string); You do need to decide whether you want all substrings (whether they overlap or not .. which your question indicates), or if you don't want to see overlapping strings. Here is a trial for such code
import java.util.Vector;
public class FindDelimitedStrings {
public static void main(String[] args) {
String[] test = getDelimitedStrings("Ron_CR7_MU'RM'_SAF_34'", "_", "'");
if (test != null) {
for (int i = 0; i < test.length; i++) {
System.out.println(" " + (i + 1) + ". |" + test[i] + "|");
}
}
}
public static String[] getDelimitedStrings(String source,
String leftDelimiter, String rightDelimiter) {
String[] answer = null;
;
Vector<String> results = new Vector<String>();
if (source == null || leftDelimiter == null || rightDelimiter == null) {
return null;
}
int loc = 0;
int begin = source.indexOf(leftDelimiter, loc);
int end;
while (begin > -1) {
end = source
.indexOf(rightDelimiter, begin + leftDelimiter.length());
if (end > -1) {
results.add(source.substring(begin, end));
// loc = end + rightDelimiter.length(); if strings must be
// returned as pairs
loc = begin + 1;
if (loc < source.length()) {
begin = source.indexOf(leftDelimiter, loc);
} else {
begin = -1;
}
} else {
begin = -1;
}
}
if (results.size() > 0) {
answer = new String[results.size()];
results.toArray(answer);
}
return answer;
}
}

How to search via character matching with a skip distance?

As the title says, I'm working on a project in which I'm searching a given text, moby dick in this case, for a key word. However instead of the word being linear, we are trying to find it via a skip distance ( instead of cat, looking for c---a---t).
I've tried multiple ways, yet can't seem to get it to actually finish one skip distance, have it not work, and call the next allowed distance (incrementing by 1 until a preset limit is reached)
The following is the current method in which this search is done, perhaps this is just something silly that I'm missing?
private int[] search()
throws IOException
{
/*
tlength is the text file length,
plength is the length of the
pattern word (cat in the original post),
text[] is a character array of the text file.
*/
int i=0, j;
int match[] = new int[2];
int skipDist = 2;
while(skipDist <= 100)
{
while(i<=tlength-(plength * skipDist))
{
j=plength-1;
while(j>=0 && pattern[j]==text[i+(j * skipDist)])j--;
if (j<0)
{
match[0] = skipDist;
match[1] = i;
return match;
}
else
{
i++;
}
}
skipDist = skipDist + 1;
}
System.out.println("There was no match!");
System.exit(0);
return match;
}

I do not know about the method you posted, but you can use this instead. I've used string and char array for this:
public boolean checkString (String s)
{
char[] check = {'c','a','t'};
int skipDistance = 2;
for(int i = 0; i< (s.length() - (skipDistance*(check.length-1))); i++)
{
boolean checkValid = true;
for(int j = 0; j<check.length; j++)
{
if(!(s.charAt(i + (j*skipDistance))==check[j]))
{
checkValid = false;
}
}
if(checkValid)
return true;
}
return false;
}
Feed the pattern to match in the char array 'check'.
String "adecrayt" evaluates true. String "cat" evaluates false.
Hope this helps.
[This part was for fixed skip distance]
+++++++++++++++++++++++++++
Now for any skip distance between 2 and 100:
public boolean checkString (String s)
{
char[] check = {'c','a','t'};
int index = 0;
int[] arr = new int[check.length];
for(int i = 0; i< (s.length()); i++)
{
if(check[index]==s.charAt(i))
{
arr[index++] = i;
}
}
boolean flag = true;
if(index==(check.length))
{
for(int i = 0; i<arr.length-1; i++)
{
int skip = arr[i+1]-arr[i];
if(!((skip>2)&&(skip<100)))
{
flag = false;
}
else
{
System.out.println("Skip Distance : "+skip);
}
}
}
else
{
flag = false;
}
return flag;
}

If you pass in a String, you only need one line:
public static String search(String s, int skipDist) {
return s.replaceAll(".*(c.{2," + skipDist + "}a.{2," + skipDist + "}t)?.*", "$1");
}
If no match found, a blank will be returned.