[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.

Have you been asked this question in an interview? Yes

思考：递归。

class Solution {
public:
    bool isScramble(string s1, string s2) {
        if(s1.size()!=s2.size()) return false;
        if(s1==s2) return true;
        //if(字符不相同) return false;
        string str1=s1;sort(str1.begin(),str1.end());
        string str2=s2;sort(str2.begin(),str2.end());
        if(str1!=str2) return false;
        int len=s1.size(),i;
        for(i=1;i<len;i++)
        {
            //if（左左&&右右)||(左右&&右左) return true;
            if((isScramble(s1.substr(0,i),s2.substr(0,i))&&isScramble(s1.substr(i),s2.substr(i)))
            ||(isScramble(s1.substr(0,i),s2.substr(len-i))&&isScramble(s1.substr(i),s2.substr(0,len-i))))
            return true;
        }
        return false;
    }
};