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题目：搅乱字符串
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
算法解析：
使用递归的思想解，
情况1：将字符串S1分割为两部分有 (0,i) (i,~) i 取值[1, len - 1], 经过若干步变换，是否能变成S2的两部分
情况2：S1 切割并且交换为两部分，然后进行若干步切割交换，最后判断两个子树是否能变成 S2 的两部分。
为了减少重复递归，使用hashmap来存储递归的结果，减少时间复杂度
代码如下：
class Solution {
public:unordered_map<string,int> recMap;bool isScramble(string s1, string s2) {if(recMap[s1+"#"+s2] == 1){return true;}if(recMap[s1+"#"+s2] == 2){return false;}int len1 = s1.length();int len2 = s2.length();if(len1 != len2){recMap[s1+"#"+s2] = 2;return false;}if(s1 == s2){recMap[s1+"#"+s2] = 1;return true;}char count[26];memset(count,0,sizeof(char)*26);for(int i = 0; i < len1; i++){count[s1[i] - 'a']++;count[s2[i] - 'a']--;}for(int i = 0; i < 26; i++){if(count[i] != 0){recMap[s1+"#"+s2] = 2;return false;}}for(int i = 1; i < len1; i++){/* 0,i i */if(isScramble(s1.substr(0,i),s2.substr(0,i)) && isScramble(s1.substr(i),s2.substr(i))){recMap[s1+"#"+s2] = 1;return true;}if(isScramble(s1.substr(i),s2.substr(0,len2 - i)) && isScramble(s1.substr(0,i),s2.substr(len2 - i))){recMap[s1+"#"+s2] = 1;return true;}}recMap[s1+"#"+s2] = 2;return false;}
};
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