[Leetcode] Scramble String

Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively.

Below is one possible representation of s1 = "great":

    great
   /    
  gr    eat
 /     /  
g   r  e   at
           / 
          a   t

To scramble the string, we may choose any non-leaf node and swap its two children.

For example, if we choose the node "gr" and swap its two children, it produces a scrambled string "rgeat".

    rgeat
   /    
  rg    eat
 /     /  
r   g  e   at
           / 
          a   t

We say that "rgeat" is a scrambled string of "great".

Similarly, if we continue to swap the children of nodes "eat" and "at", it produces a scrambled string "rgtae".

    rgtae
   /    
  rg    tae
 /     /  
r   g  ta  e
       / 
      t   a

We say that "rgtae" is a scrambled string of "great".

Given two strings s1 and s2 of the same length, determine if s2 is a scrambled string of s1.

Solution:

public class Solution {
    public boolean isScramble(String s1, String s2) {
        if(s1.equals(s2))
            return true;
        int l1=s1.length();
        int l2=s2.length();
        if(l1!=l2)
            return false;
        if(l1==0)
            return true;
        if(l1==1)
            return s1.equals(s2);
        char[] c_s1=s1.toCharArray();
        char[] c_s2=s2.toCharArray();
        Arrays.sort(c_s1);
        Arrays.sort(c_s2);
        for(int i=0;i<c_s1.length;++i){
            if(c_s1[i]!=c_s2[i])
                return false;
        }
        boolean b=false;
        for(int i=1;i<l1&&!b;++i){
            String s11=s1.substring(0,i);
            String s12=s1.substring(i);
            String s21=s2.substring(0,i);
            String s22=s2.substring(i);
            b=isScramble(s11, s21)&&isScramble(s12, s22);
            if(!b){
                String s31=s2.substring(0, l1-i);
                String s32=s2.substring(l1-i);
                b=isScramble(s11, s32)&&isScramble(s12, s31);
            }
        }
        return b;
    }
}

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20150220:

public class Solution {
    public boolean isScramble(String s1, String s2) {
        if(s1==null||s2==null||s1.length()!=s2.length())
            return false;
        if(s1.equals(s2))
            return true;
        char[] c1=s1.toCharArray();
        char[] c2=s2.toCharArray();
        Arrays.sort(c1);
        Arrays.sort(c2);
        for(int i=0;i<c1.length;++i){
            if(c1[i]!=c2[i])
                return false;
        }
        int N=s1.length();
        for(int i=1;i<N;++i){
            String s1_temp1=s1.substring(0,i);
            String s1_temp2=s1.substring(i);
            String s2_temp1=s2.substring(0,i);
            String s2_temp2=s2.substring(i);
            if(isScramble(s1_temp1, s2_temp1)&&isScramble(s1_temp2, s2_temp2))
                return true;
            String s2_temp3=s2.substring(0,N-i);
            String s2_temp4=s2.substring(N-i);
            if(isScramble(s1_temp1, s2_temp4)&&isScramble(s1_temp2, s2_temp3))
                return true;
        }
        return false;
    }
}