[leedcode 87] 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.

public class Solution {
    public boolean isScramble(String s1, String s2) {
    /*  首先选择递归，简单明了，对两个string进行partition，然后比较四个字符串段。但是递归的话，这个时间复杂度比较高。然后想到能否DP，但是           即使用DP的话，也要O(n^3)。还是在递归里做些剪枝，这样就可以避免冗余计算：
        对于每两个要比较的partition，需要统计他们字符出现次数（字符个数和种类要一致），如果不相等返回。*/
        int s1Len=s1.length();
        int s2Len=s2.length();
        if(s1Len!=s2Len) return false;
        int frequent[]=new int[26];
        for(int i=0;i<s1Len;i++){
            int temp=s1.charAt(i)-'a';
            frequent[temp]++;
        }
        for(int i=0;i<s2Len;i++){
            int temp=s2.charAt(i)-'a';
            frequent[temp]--;
        }
        for(int i=0;i<frequent.length;i++){
            if(frequent[i]!=0)return false;
        }
        if(s1Len==1&&s2Len==1) return true;/////
        
/*        报错：Line 26: java.lang.StackOverflowError   
        if(s1.equals(s2)) return true;
        char[] s11=s1.toCharArray();
        char[] s22=s2.toCharArray();
        Arrays.sort(s11);
        Arrays.sort(s22);
        
        if(!(String.valueOf(s11)).equals(String.valueOf(s22))) return false;*/
        
        for(int i=0;i<s1Len;i++){
            boolean temp=isScramble(s1.substring(0,i),s2.substring(0,i))&&isScramble(s1.substring(i),s2.substring(i));
            if(temp) return true;
            temp=isScramble(s1.substring(0,i),s2.substring(s2.length()-i))&&isScramble(s1.substring(i),s2.substring(0,s2.length()-i));
            if(temp) return true;
        }
        return false;
    }
}