leetcode scramble-string

问题描述：

https://oj.leetcode.com/problems/scramble-string/点击打开链接

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.

问题分析：

1. str1, str2 长度相等，字符种类一样才可能满足条件。

2. str1,str2 有两种切割方法  

• str1[0, i) str1[i, n) str2[0, i), str2[i, n)
• str1[0, i) str1[i, n) str2[n-i, i), str2[0, n-i)

代码示例：

class Solution {public:    bool isScramble(string s1, string s2)     {        string s3(s1), s4(s2);        sort(s3.begin(), s3.end());        sort(s4.begin(), s4.end());        if (s3 != s4) { return false; }        int n = s3.length();        if (n == 1) { return true; }                for (int i = 1; i < n; i++)        {            bool result = isScramble(s1.substr(0, i), s2.substr(0, i)) && isScramble(s1.substr(i, n - i), s2.substr(i, n - i));            result = result || isScramble(s1.substr(0, i), s2.substr(n - i, i)) && isScramble(s1.substr(i, n - i), s2.substr(0, n - i));            if (result) { return true; }        }        return false;    }};