87. Scramble String

题目：

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.

代码：

class Solution {public:    bool isScramble(string s1, string s2) {        if(s1 == s2)return true;        if(s1.size() != s2.size())return false;        vector<int> abc(26, 0);        for(int i = 0; i < s1.size(); i++)        {            abc[s1[i] - 'a']++;            abc[s2[i] - 'a']--;        }        for(auto p : abc) if(p != 0) return false;/*这个判断非常重要，有人说给的s1，s2一定包含相同字母，但是当递归到子串的时候就不行了。*/        for(int i = 1; i < s1.size(); i++)        {            if(isScramble(s1.substr(0, i), s2.substr(0, i)) && isScramble(s1.substr(i), s2.substr(i))            || isScramble(s1.substr(0, i), s2.substr(s1.size() - i)) && isScramble(s1.substr(i), s2.substr(0, s1.size() - i) ))            return true;        }        return false;    }};