卖萌的弱渣

I am stupid, I am hungry.

Scramble String

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”:

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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”.

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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”.

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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.

Solution

s1和s2是scramble的话，那么必然存在一个在s1上的长度l1，将s1分成s11和s12两段，同样有s21和s22。 * 那么要么s11和s21是scramble的并且s12和s22是scramble的 * 要么s11和s22是scramble的并且s12和s21是scramble的

(Scramble-String.py) download
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class Solution(object):
    def isScramble(self, s1, s2):
        """
        :type s1: str
        :type s2: str
        :rtype: bool
        """

        if s1 == s2:
            return True

        if len(s1) != len(s2):
            return False

        # 把s1,s2转换成数组
        l1 = list(s1)
        l2 = list(s2)
        l1 = sorted(l1)
        l2 = sorted(l2)

        if l1 != l2:
            return False

        length = len(l1)
        for i in range(1,length):
            # 没有交换s11==s21 and s12==s22
            if self.isScramble(s1[i:],s2[i:]) and self.isScramble(s1[:i],s2[:i]):
                return True
            # 有交换 s11==s22 and s12 == s21
            if self.isScramble(s1[i:],s2[:length-i]) and self.isScramble(s1[:i], s2[length-i:]):
                return True
        return False