[leetcode]Scramble String @ Python
原题地址：https://oj.leetcode.com/problems/scramble-string/
题意：
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.
解题思路：二叉树的问题一般使用递归来解决。
代码：
class Solution:# @return a boolean
def isScramble(self, s1, s2):
if len(s1)!=len(s2): return False
if s1==s2: return True
l1=list(s1); l2=list(s2)
l1.sort();l2.sort()
if l1!=l2: return False
length=len(s1)
for i in range(1,length):
if self.isScramble(s1[:i],s2[:i]) and self.isScramble(s1[i:],s2[i:]): return True
if self.isScramble(s1[:i],s2[length-i:]) and self.isScramble(s1[i:],s2[:length-i]): return True
return False
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