# 87. Scramble String

**<font color=red>难度:Hard<font>**

## 刷题内容
> 原题连接

* https://leetcode.com/problems/scramble-string/

> 内容描述

```
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.

Example 1:

Input: s1 = "great", s2 = "rgeat"
Output: true
Example 2:

Input: s1 = "abcde", s2 = "caebd"
Output: false
```

> 思路1
******- 时间复杂度: O(2^n)******- 空间复杂度: O(1)******

刚开始的时候题目的意思没理解清楚，以为是把字符串对半，其实字符串有很多的二叉树表示方法。第一种方法就是递归的方法。每次递归时比较i到len的之间的子串是否和s2中的子串的相同，这里有两种比较方式，s2开始和结尾两种比较方式，有一种满足即可，这里我们只要判断两个子串的字母是否相同。相同就进行下一次递归。

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

> 思路2
******- 时间复杂度: O(n^4)******- 空间复杂度: O(n^3

第二种的方法使用DP，定义数组dp[i][j][t],表示s1[i]和s2[j]开始的长度为t的字符是否为scramble string。

```cpp
class Solution {
public:
    bool isScramble(string s1, string s2) {
    int len = s1.length();
    int dp[len][len][len + 1] = {0};
    for(int i = len - 1;i >= 0;--i)
        for(int j = len - 1;j >= 0;--j)
            for(int t = 1;t <= min(len - i,len - j);++t)
                if(t == 1)
                    dp[i][j][t] = (s1[i] == s2[j]);
                else
                {
                    int k = 1;
                    for(;k < t;++k)
                        if((dp[i][j][k] && dp[i + k][j + k][t - k]) || (dp[i][j + t - k][k] && dp[i + k][j][t - k]))
                            break;
                    dp[i][j][t] = k == t ? 0 : 1;
                }
    return dp[0][0][len];

}
};
```