难度: Hard
原题连接
内容描述
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
思路 1 - 时间复杂度: O(N^2lgN)- 空间复杂度: O(1)******
递归，很简单
beats 84.62%
class Solution(object):
def isScramble(self, s1, s2):
"""
:type s1: str
:type s2: str
:rtype: bool
"""
if len(s1) != len(s2) or sorted(s1) != sorted(s2):
return False
if len(s1) < 4 or s1 == s2:
return True
return any(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]) for i in range(1, len(s1)))思路 2 - 时间复杂度: O(N^2lgN)- 空间复杂度: O(N)******
加一下cache，beats 97.29%
class Solution(object):
def __init__(self):
self.cache = {}
def isScramble(self, s1, s2):
"""
:type s1: str
:type s2: str
:rtype: bool
"""
if (s1, s2) in self.cache:
return self.cache[(s1, s2)]
if len(s1) != len(s2) or sorted(s1) != sorted(s2):
self.cache[(s1, s2)] = False
return False
if len(s1) < 4 or s1 == s2:
self.cache[(s1, s2)] = True
return True
for i in range(1, len(s1)):
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
self.cache[(s1, s2)] = False
return False