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.

Example 1:

Input: s1 = "great", s2 = "rgeat"
Output: true
Example 2:

Input: s1 = "abcde", s2 = "caebd"
Output: false

简单的说，就是s1和s2是scramble的话，那么必然存在一个在s1上的长度l1，将s1分成s11和s12两段，同样有s21和s22。
那么要么s11和s21是scramble的并且s12和s22是scramble的；
要么s11和s22是scramble的并且s12和s21是scramble的。

两个字符串的相似的必备条件是含有相同的字符集。简单的做法是把两个字符串的字符排序后，然后比较是否相同。
加上这个检查就可以大大的减少递归次数。

class Solution {
    public boolean isScramble(String s1, String s2) {
        if(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);
        if(!Arrays.equals(c1, c2)){
            return false;
        }
        
        for(int i=1; i<s1.length(); i++){
            String s11 = s1.substring(0,i);
            String s12 = s1.substring(i);
            
            String s21 = s2.substring(0,i);
            String s22 = s2.substring(i);
            
            if(isScramble(s11, s21) && isScramble(s12, s22)) return true;
           
            s21 = s2.substring(0, s1.length()-i);
            s22 = s2.substring(s1.length()-i);
            if(isScramble(s11, s22) && isScramble(s12, s21)) return true;
        }
        return false;
    }
}