[Leetcode]87. 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
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
