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
– azurefrog, Commented May 24, 2017 at 18:48
if i didnt use break statement, then op will be 12141618343638565878 . — Prashant Gaurav
– Prashant Gaurav, Commented May 24, 2017 at 18:49
Does output need to be sorted? Or interleaved? It's unclear. — Jorn Vernee
– Jorn Vernee, Commented May 24, 2017 at 18:58
@PrashantGaurav If s1="97531" and s2="2468", what output would you expect? 123456789 or 927456381? — azurefrog
– azurefrog, Commented May 24, 2017 at 19:02

Roman Puchkovskiy · Accepted Answer · 2017-05-24 18:50:51Z

3

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.

answered May 24, 2017 at 18:50

Roman Puchkovskiy

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Comments

Jorn Vernee · Accepted Answer · 2017-05-24 18:55:44Z

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, 2017 at 18:55

Jorn Vernee

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3 Comments

azurefrog Over a year ago

But that only works if the inputs are strictly sorted. Can we really assume that's always the case?

Jorn Vernee Over a year ago

@azurefrog I was just wondering about that. tbh I thought that was the whole point, to get sorted output.

azurefrog Over a year ago

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.

Yahya · Accepted Answer · 2017-05-24 18:56:14Z

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, 2017 at 18:56

Yahya

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3 Comments

azurefrog Over a year ago

But that only works if the inputs are strictly sorted. Can we really assume that's always the case?

Yahya Over a year ago

@azurefrog What do you mean by strictly sorted?, I used the example the OP mentioned.

azurefrog Over a year ago

True, see my reply on @Jorn Vernee's answer. The OP's question is ambiguous.

Tezra · Accepted Answer · 2017-05-24 18:58:28Z

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, 2017 at 18:58

answered May 24, 2017 at 18:53

Tezra

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2 Comments

Chris Gilardi Over a year ago

This might be a little redundant with the double checking of s1.length() and s2.length().

Tezra Over a year ago

@ChrisGilardi true... Broke out that bit so that the while loop does minimal work on while part.

Chris Gilardi · Accepted Answer · 2017-05-24 18:55:12Z

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, 2017 at 18:55

Chris Gilardi

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5 Comments

Roman Puchkovskiy Over a year ago

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.

Chris Gilardi Over a year ago

@RomanPuchkovskiy thanks for the insight, I wasn't aware that this was the case. However, I did just read in this answer that it may not matter, because Java's compiler is smart enough to turn it to a StringBuilder. Either way, thanks!

dumbPotato21 Over a year ago

@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.

Chris Gilardi Over a year ago

@ChandlerBing Well alrighty, I stand corrected! Thanks for letting me know, I'll remember that for my own projects!

dumbPotato21 Over a year ago

@ChrisGilardi No worries :) I also recommend reading this stackoverflow.com/a/47758/5055190

swapnil · Accepted Answer · 2017-07-13 05:26:42Z

0

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++;
    }
}

}

edited Jul 13, 2017 at 5:26

answered May 24, 2017 at 19:06

swapnil

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1 Comment

Roman Puchkovskiy Over a year ago

For String, StringBuffer, and StringBuilder, charAt() is a constant-time operation. stackoverflow.com/questions/6461402/…

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