87. Scramble String Leetcode Python

最新推荐文章于 2025-07-03 11:30:19 发布

hyperbolechi

最新推荐文章于 2025-07-03 11:30:19 发布

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分类专栏： leetcode 文章标签： Backtracking python leetcode

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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.

This problem can be solved with brute Force.

class Solution:
    # @param s1, a string
    # @param s2, a string
    # @return a boolean
    def isScramble(self, s1, s2):
        n, m = len(s1),len(s2)
        if n!=m or sorted(s1)!=sorted(s2):
            return False
        if n < 4 or s1 == s2:
            return True
        for i in range(1, n):
            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