LeetCode-87-Scramble String

算法描述：

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

解题思路：递归。每次计算当前分割下的字符数，如果不相等，则返回假，否则，返回真。

    bool isScramble(string s1, string s2) {
        if(s1==s2) return true;
        vector<int> count(26,0);
        
        for(char c:s1){
            count[c-'a']++;
        }
        for(char c:s2){
            count[c-'a']--;
        }
        for(int i=0; i < 26; i++){
            if(count[i]<0) return false;
        }
        int len = s1.size();
        for(int i =1; i < s1.size(); i++){
            if(isScramble(s1.substr(0,i),s2.substr(0,i)) && isScramble(s1.substr(i,len-i),s2.substr(i,len-i))) return true;
            if(isScramble(s1.substr(0,i),s2.substr(len-i,i))&& isScramble(s1.substr(i,len-i),s2.substr(0,len-i))) return true;
        }
        return false;
    }