20170321-leetcode-131-Palindrome partitioning

1.Description

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"]]

https://leetcode.com/problems/palindrome-partitioning/#/description
解读
输入：一个字符串，把该字符串分割，使得每个字符串都构成回文，输出所有满足条件的分割结果
求解该问题，可以采取回溯和循环的策略。
如果输入“aab”,循环判断[0:1],[0:2],[0:3] (左开右闭)，对于其中的每一个采取深度优先搜索（DFS)

https://leetcode.com/problems/palindrome-partitioning/#/solutions
如上图，对于[0:1]—>分割成[a]+[a][a][b]/[a][ab]
对于[0:2]—>分割成[aa]+[b]
对于[0:3]—>分割成[aab]

2.Solution

class Solution(object):    def partition(self, s):        return [[s[:i]] + rest                for i in range(1, len(s) + 1)                if s[:i] == s[i - 1::-1]                for rest in self.partition(s[i:])] or [[]]

https://leetcode.com/problems/palindrome-partitioning/#/solutions
if s[:i] == s[i - 1::-1]，s[i-1::-1]是s[:i]的逆序，判断是否为回文
上述代码可以修改为：

class Solution(object):    def mypartition(self, s):        ans = []        for i in range(1, len(s) + 1):            if s[:i] == s[i - 1::-1]:                for rest in self.partition(s[i:]):                    ans.append([s[:i]] + rest)                    print(ans)        if not ans:            return [[]]        return ans