87. Scramble String

1. 问题描述
   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.

2. 解决思路
   递归解决，每次分成两个部分，看是否能满途条件。（注意，需要在开头判断条件是否符合，不符合马上return false，不然时间复杂度很高）

3. 代码

class Solution {public:    bool isScramble(string s1, string s2) {        if (s1 == s2)            return true;        int count[26] = {0};        for (int i = 0; i < s1.length(); ++i) {            ++count[s1[i]-'a'];            --count[s2[i]-'a'];        }        for (int i = 0; i < 26; ++i)            if (count[i] != 0)                return false;        int len = s1.length();        for (int i = 1; i < len; ++i ) {            if (isScramble(s1.substr(0,i),s2.substr(0,i)) && isScramble(s1.substr(i),s2.substr(i)))                return true;            else if (isScramble(s1.substr(0,i),s2.substr(len-i)) && isScramble(s1.substr(i),s2.substr(0,len-i)))                return true;        }        return false;    }};