給出兩個(gè)字符串，可以進(jìn)行任意多次操作，每次將字符串切分成兩個(gè)，選擇對(duì)調(diào)兩個(gè)子串or不對(duì)調(diào)，問(wèn)字符串s1能否經(jīng)過(guò)操作變?yōu)樽址?br />
用了比較暴力的做法，窮舉所有操作的可能性，用了Discussion中記錄下已經(jīng)暴力枚舉過(guò)的結(jié)果，否則會(huì)TLE
https://leetcode.com/problems/scramble-string/solutions/3357439/easy-solutions-in-java-python-and-c-look-at-once-with-exaplanation
1 #1402
2 #Runtime: 222 ms (Beats 21.95%)
3 #Memory: 20.5 MB (Beats 7.32%)
4
5 class Solution(object):
6 def isScramble(self, s1, s2):
7 """
8 :type s1: str
9 :type s2: str
10 :rtype: bool
11 """
12 if len(s1) != len(s2):
13 return False
14 if s1 == s2:
15 return True
16 if len(s1) == 1:
17 return False
18 t = s1 + " " + s2
19 if t in self.solved:
20 return self.solved[t]
21 for i in range(1, len(s1)):
22 if self.isScramble(s1[0:i], s2[0:i]) and self.isScramble(s1[i:], s2[i:]):
23 self.solved[t] = True
24 return True
25 if self.isScramble(s1[0:i], s2[len(s1)-i:]) and self.isScramble(s1[i:], s2[0:len(s1)-i]):
26 self.solved[t] = True
27 return True
28 self.solved[t] = False
29 return False
30 solved = {}