How to merge two String into one String in java

Question

Suppose:

String s1="13579";
String s2="2468";

then output would be 123456789

Here my code :

public class JoinString {

    public static void main(String[] args) {

        String s1 = "13579"; 
        String s2 = "2468";
        for (int i = 0; i < s1.length(); i++) {
            for (int j = i; j < s2.length(); j++) {

                System.out.print(s1.charAt(i) + "" + s2.charAt(j));
                break;
            }
        }
    }
}

You don't really need two nested loops there. Just one loop that keeps track of an index into both strings is sufficient. Once you've run out of the shorter one, you can just append a substring of the longer one based on how far into it you've gotten. — azurefrog, May 24 '17 at 18:48
if i didnt use break statement, then op will be 12141618343638565878 . — Prashant Gaurav, May 24 '17 at 18:49
@RïshïKêshKümar won't work, since the OP is trying to interweave the strings, not append one to the other. — azurefrog, May 24 '17 at 18:50
@RïshïKêshKümar, according to you logic. o/p will be 135792468 — Prashant Gaurav, May 24 '17 at 18:52
@azurefrog, can you show me logic. as in above code. it is not printing last char of s1 i.e 9. — Prashant Gaurav, May 24 '17 at 18:53
@PrashantGaurav Go look at Roman's answer. It doesn't even need the fix after the loop, since it's clever with the termination condition. — azurefrog, May 24 '17 at 18:54
Does output need to be sorted? Or interleaved? It's unclear. — Jorn Vernee, May 24 '17 at 18:58
@PrashantGaurav If `s1="97531"` and `s2="2468"`, what output would you expect? `123456789` or `927456381`? — azurefrog, May 24 '17 at 19:02

score 3 · Answer 1 · answered May 24 '17 at 18:50

StringBuilder buf = new StringBuilder();
for (int i = 0; i < Math.max(s1.length(), s2.length()); i++) {
    if (i < s1.length()) {
        buf.append(s1.charAt(i));
    }
    if (i < s2.length()) {
        buf.append(s2.charAt(i));
    }
}
final String result = buf.toString();

You only need one loop. Also, you can use a StringBuilder class to build your string character by character.

score 3 · Answer 2 · answered May 24 '17 at 18:55

3

How about this little trick:

String s1 = "13579"; 
String s2 = "2468";

String s3 = (s1 + s2).codePoints() // stream of code points
    .sorted()
    // collect into StringBuilder
    .collect(StringBuilder::new, StringBuilder::appendCodePoint,  StringBuilder::append) 
    .toString();

System.out.println(s3); // 123456789

answered May 24 '17 at 18:55

Jorn Vernee

31,735
4
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But that only works if the inputs are strictly sorted. Can we really assume that's always the case? – azurefrog May 24 '17 at 18:56
@azurefrog I was just wondering about that. tbh I thought that was the whole point, to get sorted output. – Jorn Vernee May 24 '17 at 18:57
Hmm, good point, looking back at it, the OP's question could be read as "every other character" or "sorted output", it's sort of ambiguous. – azurefrog May 24 '17 at 18:58

score 2 · Answer 3 · answered May 24 '17 at 18:56

2

A simple way to achieve that is by doing something like this:

String s1 = "13579"; 
String s2 = "2468";
char[]result = (s1+s2).toCharArray();
Arrays.sort(result);
System.out.println(result);

Output:

123456789

answered May 24 '17 at 18:56

Yahya

13,349
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But that only works if the inputs are strictly sorted. Can we really assume that's always the case? – azurefrog May 24 '17 at 18:57
@azurefrog What do you mean by *strictly sorted*?, I used the example the OP mentioned. – Yahya May 24 '17 at 18:59
True, see my reply on @Jorn Vernee's answer. The OP's question is ambiguous. – azurefrog May 24 '17 at 19:00

Tezra · Answer 4 · 2017-05-24T18:58:28.270

1

You just need to combine the logic of the two loops like this (and use StringBuilder if you want to build a string this way).

    String s1 = "13579"; 
    String s2 = "2468";
    int length = Math.max(s1.length(), s2.length());
    for (int i = 0; i < length; i++) {
        if(i < s1.length())
          System.out.print(s1.charAt(i));

        if(i < s2.length())
          System.out.print(s2.charAt(i));
    }

edited May 24 '17 at 18:58

answered May 24 '17 at 18:53

Tezra

8,463
3
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This might be a little redundant with the double checking of `s1.length()` and `s2.length()`. – Chris Gilardi May 24 '17 at 18:56
@ChrisGilardi true... Broke out that bit so that the while loop does minimal work on while part. – Tezra May 24 '17 at 18:59

score 0 · Answer 5 · answered May 24 '17 at 18:55

0

Similar to @Roman Puchovskiy, here is a way to do it without the StringBuilder.

    String s1 = "abcde";
    String s2 = "defgh";
    String combined = "";
    int length = s1.length() > s2.length() ? s1.length() : s2.length(); //Finds the longest length of the two, to ensure no chars go missing
    for(int i = 0 ; i < length ; i++) {
        if(i < s1.length()) { // Make sure there is a char in s1 where we're looking.
            combined += s1.charAt(i);
        }
        if(i < s2.length()) { // Same with s2.
            combined += s2.charAt(i);
        }
    }

Where combined becomes the combined string. ("adbecfdgeh").

Hope this helps!

answered May 24 '17 at 18:55

Chris Gilardi

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Please note that this will produce a lot of garbage: a string object per `+=`, so an object per resulting string character. A GC may not be happy. – Roman Puchkovskiy May 24 '17 at 18:59
@RomanPuchkovskiy thanks for the insight, I wasn't aware that this was the case. However, I did just read in [this answer](https://stackoverflow.com/a/4323132/4914803) that it may not matter, because Java's compiler is smart enough to turn it to a `StringBuilder`. Either way, thanks! – Chris Gilardi May 24 '17 at 19:04
2

@ChrisGilardi the compiler does do that, but then `toString` is called each time the loop runs, while with a `StringBuilder`, that happens only once. May not be noticiable here, but this does take more memory than a `StringBuilder` solution. For single statement, it doesn't matter. But within loops, it makes a considerable difference. – dumbPotato21 May 24 '17 at 19:06
@ChandlerBing Well alrighty, I stand corrected! Thanks for letting me know, I'll remember that for my own projects! – Chris Gilardi May 24 '17 at 19:07
@ChrisGilardi No worries :) I also recommend reading this https://stackoverflow.com/a/47758/5055190 – dumbPotato21 May 24 '17 at 19:09

swapnil · Answer 6 · 2017-07-13T05:26:42.867

Similar way as merge method in merge sort.

Also it's better if you convert string to char array so that random access to element is in constant time. Because charAt has O(n) time complexity.

public class JoinString {

public static void main( String[] args ) {

    String s1 = "13579";
    String s2 = "2468";
    final char[] str1 = s1.toCharArray();
    final char[] str2 = s2.toCharArray();
    int i;
    for ( i = 0; i < str1.length && i < str2.length; i++ ) {
        System.out.print( str1[i] + "" + str2[i] );
    }

    while ( i < str1.length ) {
        System.out.print( str1[i] + "" );
        i++;
    }

    while ( i < str2.length ) {
        System.out.print( str2[i] + "" );
        i++;
    }
}

}

For `String`, `StringBuffer`, and `StringBuilder`, `charAt()` is a constant-time operation. https://stackoverflow.com/questions/6461402/java-charat-and-deletecharat-performance — Roman Puchkovskiy, May 25 '17 at 05:29

How to merge two String into one String in java

6 Answers6