136. Palindrome Partitioning / 131. Palindrome Partitioning

• 本題難度: Medium

• Topic: Search DFS

  Description

  Given a string s, partition s such that every substring of the partition is a palindrome.

Return all possible palindrome partitioning of s.

Example
Given s = "aab", return:

[
["aa","b"],
["a","a","b"]]

我的代碼

recursion

class Solution:"""@param: s: A string@return: A list of lists of string"""def partition(self, s):# write your code heredef dfs(s,part, res):if s == '':res.append(part)return for i in range(1,len(s)+1):if isValid(s[:i]):dfs(s[i:],part+[s[:i]],res)def isValid(s):l = len(s)for i in range(l//2):if s[i]!=s[l-1-i]:return Falsereturn Trueres = []dfs(s, [], res)return res

思路
非常常見的DFS.
res為全局變量。所以不需要return什么值

轉載于:https://www.cnblogs.com/siriusli/p/10392217.html

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