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# Q86. Scramble String

直达: <https://leetcode.com/problems/scramble-string/description/>

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.

## 分析

将s1分成两个子字符串s11和s12, 则无论怎么变化, s11和s12的所有的字符串都在s2的相同的子树里，或者左子树s21，或者右子树s12。

如果s11与s21并且s12与s22都满足Scramble或者s11与s22并且s12与s21都满足Scramble，则s1与s2满足Scramble。显然，需要通过递归求解。

## C++代码

```cpp
class Solution {
public:
    bool isScramble(string s1, string s2) {
        if(s1 == s2) return true;
        if(s1.size() != s2.size()) return false;
        string str1 = s1; string str2 = s2;
        sort(str1.begin(), str1.end());
        sort(str2.begin(), str2.end());
        if (str1 != str2) return false;
        for(int i = 1; i < str1.size(); i++){
            string s11 = s1.substr(0, i);
            string s12 = s1.substr(i);
            string s21 = s2.substr(0,i);
            string s22 = s2.substr(i);
            if(isScramble(s11, s21) && isScramble(s12, s22)) return true;
            s21 = s2.substr(s1.size()-i);
            s22 = s2.substr(0, s1.size()-i);
            if(isScramble(s11, s21) && isScramble(s12, s22)) return true;
        }
        return false;
    } 
};
```