如何将char数组存储到字符串数组中并在java中对该字符串数组进行排序？

Question

Given 2 strings A and B, print all the interleavings of the 2 strings. An interleaved string of given two strings preserves the order of characters in individual strings and uses all the characters of both the strings. For simplicity, you can assume that the strings have unique characters.

input:
2
nkb gl
bn zh
i wrote some code but im not getting the sorted strings.

public class Test {

static void interleavings(String A, String B, char[] ans, int m, int n, int idx) 
{
    if(m==0 && n==0) 
    {
        System.out.println(ans);
        return;
    } 
    if(m != 0) 
    {
        ans[idx]=A.charAt(0);
        interleavings(A.substring(1,m), B, ans, m-1, n, idx+1);
    } 
    if(n != 0) 
    {
        ans[idx]= B.charAt(0);
        interleavings(A, B.substring(1,n), ans, m, n-1, idx+1);
    }
}
public static void main(String[] args) 
{
    Scanner s =  new Scanner(System.in);
    int t = s.nextInt();
    for(int i = 0; i < t; i++)
    {
        String s1=s.next();
        String s2=s.next();
        char [] ans = new char[s1.length()+s2.length()];
        System.out.println("Case #"+(i+1)+":");
        interleavings(s1,s2,ans,s1.length(),s2.length(),0);  
    }
}

}

My output
Case #1:
nkbgl
nkgbl
nkglb
ngkbl
ngklb
nglkb
gnkbl
gnklb
gnlkb
glnkb
Case #2:
bnzh
bznh
bzhn
zbnh
zbhn
zhbn
Expected output
Case #1:
glnkb
gnkbl
gnklb
gnlkb
ngkbl
ngklb
nglkb
nkbgl
nkgbl
nkglb
Case #2:
bnzh
bzhn
bznh
zbhn
zbnh
zhbn

I'm new to java can someone please guide me. how to over come this problem please.

Answer 1

I would no use char array as it will make things more complicated. I would pass a current value as string. See curr + A.charAt(0) this will append the char to a String , this way you are not dealing with char .

Store results in Collection . Than you can sort the whole collection an print it when done, instead of doing it on the fly.

class Main {

    public static void interleavings(String curr, String A, String B, Set<String> result) {
        if (A.length() == 0 && B.length() == 0) {
            result.add(curr);
            return;
        }
        if (A.length() > 0) {
            interleavings(curr + A.charAt(0), A.substring(1), B, result);
        }
        if (B.length() > 0) {
            interleavings(curr + B.charAt(0), A, B.substring(1), result);
        }
    }

    public static void main(String[] args) {
        String A = "bn";
        String B = "zh";

        Set<String> result = new HashSet<>();
        interleavings("", A, B, result);

        // this will print all the results sorted
        result.stream()
              .sorted()
              .forEach(System.out::println);
    }
}

Answer 2

You should swap a and b when a.compareTo(b) > 0 and then make recursive calls.

static void interleavings(String a, String b, String result) {
    if (a.isEmpty() && b.isEmpty())
        System.out.println(result);
    else if (a.compareTo(b) > 0)
        interleavings(b, a, result);
    else {
        if (!a.isEmpty())
            interleavings(a.substring(1), b, result + a.charAt(0));
        if (!b.isEmpty())
            interleavings(a, b.substring(1), result + b.charAt(0));
    }
}

and

interleavings("nkb", "gl", "");
interleavings("bn", "zh", "");

output

glnkb
gnkbl
gnklb
gnlkb
ngkbl
ngklb
nglkb
nkbgl
nkgbl
nkglb
bnzh
bzhn
bznh
zbhn
zbnh
zhbn