标签：ret ssi span append key return style false []
Given a string s, partition s such that every substring of the partition is a palindrome.
Return all possible palindrome partitioning of s.
For example, given s = "aab",
Return
[ ["aa","b"], ["a","a","b"] ]
Keys: 1. 一般列出所有组合，可能性，etc 的题目都可以用DFS.
2. L8. 对于Class Solution中的全局变量x，可以用Solution.x
1 class Solution(object): 2 def partition(self, s): 3 """ 4 :type s: str 5 :rtype: List[List[str]] 6 """ 7 Solution.res = [] 8 self.DFS(s, []) 9 return Solution.res 10 11 def DFS(self, s, stringlist): 12 if len(s) == 0: 13 Solution.res.append(stringlist) 14 # 如果一开始的substring是palin，把他存到stringlist中 15 for i in range(1, len(s)+1): 16 if self.isPalin(s[:i]): 17 self.DFS(s[i:], stringlist + [s[:i]]) 18 19 20 def isPalin(self, s): 21 for i in range(len(s)): 22 if s[i] != s[len(s)-1-i]: 23 return False 24 return True 25 26
Leetcode 131. Palindrome Partitioning
标签：ret ssi span append key return style false []
原文地址：http://www.cnblogs.com/lettuan/p/6079806.html
