87. Scramble String Leetcode Python

标签：backtracking   python   leetcode   

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
        
            

87. Scramble String Leetcode Python

标签：backtracking   python   leetcode   

原文地址：http://blog.csdn.net/hyperbolechi/article/details/45082387