[Leetcode]87. Scramble String

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[[Leetcode]87. Scramble String](https://leetcode.com/problems/scramble-string/)

- 本题难度: Hard
- Topic: divide and conquere
# 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.

Example 1:

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

Example 2:

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

# 我的代码 
```python

class Solution:
    
    def isScramble(self, s1: str, s2: str) -> bool:
        l = len(s1)
        if l!=len(s2):
            return False
        if s1 == s2:
            return True
        if sorted(s1)!=sorted(s2):
            return False
        for i in range(1,l):
            if (self.isScramble(s1[:i],s2[:i]) and self.isScramble(s1[i:],s2[i:])) or (self.isScramble(s1[:i],s2[-i:]) and self.isScramble(s1[i:],s2[:-i])):
                return True
        return False

```

 

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