leetcode之Scramble String

原题如下：

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.

这道题目还是有一定难度的，借鉴大牛的博客，下面说一下递归求解的思路：

1）如果两个字符串长度不等，则返回false

2）如果两个字符串相等，则返回true

3）将两个字符串排序后比较，如果不相等则返回false

4）将s1分割成左右子树两部分，同时将s2分割成左右子树两部分，这样就可以划分成两个子问题进行求解，这里要注意的是还需要交叉验证，只要有一个满足就返回true

5）遍历结束都没有合适的分割点返回false

class Solution {public:    bool isScramble(string s1, string s2) {if(s1.size() != s2.size())return false;if(s1 == s2)return true;string str1 = s1,str2 = s2;sort(str1.begin(),str1.end());sort(str2.begin(),str2.end());if(str1 != str2)return false;int len = s1.size();for(int i = 1; i < len ; i++){string leftS1 = s1.substr(0,i);string rightS1 = s1.substr(i);string leftS2 = s2.substr(0,i);string rightS2 = s2.substr(i);if(isScramble(leftS1,leftS2) && isScramble(rightS1,rightS2))return true;leftS2 = s2.substr(0,len - i);rightS2 = s2.substr(len - i);if(isScramble(leftS1,rightS2) && isScramble(rightS1,leftS2))return true;}return false;    }};