leetcode 131. Palindrome Partitioning（python）

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描述

Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s.

A palindrome string is a string that reads the same backward as forward.

Example 1:

Input: s = "aab"
Output: [["a","a","b"],["aa","b"]]
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Example 2:

Input: s = "a"
Output: [["a"]]
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Note:

1 <= s.length <= 16
s contains only lowercase English letters.
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解析

根据题意，给定一个字符串 s ，对 s 分区使得分区的每个子串都是一个回文。 返回 s 的所有可能的回文分区，回文字符串就是向后读取与向前读取相同的字符串。

这道题是典型的 DFS + 动态规划的题，非常经典。因为这道题要找出所有的组合，所以肯定是要暴力搜索的，所以先用动态规划数组 dp[i][j] 保存 s[i][j] 是否为回文字符串。然后使用 DFS 找出所有的组合即可。

解答

import numpy as np
class Solution(object):
    def __init__(self):
        self.dp = None
        self.result = []
    def partition(self, s):
        """
        :type s: str
        :rtype: List[List[str]]
        """
        N = len(s)
        self.dp = [[False]*N for _ in range(N)]
        for i in range(N):
            self.dp[i][i] = True
        for i in range(N-1):
            self.dp[i][i+1] = (s[i] == s[i+1])
        for L in range(3, N+1):
            for i in range(N-L+1):
                j = i+L-1
                if s[i]==s[j] and self.dp[i+1][j-1]:
                    self.dp[i][j] = True
        self.dfs(0, [], s, N)
        return self.result
    
    def dfs(self, i, t, s, N):
        if i == N:
            self.result.append(t)
            return 
        for j in range(i, N):
            if self.dp[i][j]:
                self.dfs(j+1, t+[s[i:j+1]], s, N)
        
      	      
		
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运行结果

Runtime: 668 ms, faster than 49.52% of Python online submissions for Palindrome Partitioning.
Memory Usage: 39.2 MB, less than 5.03% of Python online submissions for Palindrome Partitioning.
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原题链接：leetcode.com/problems/pa…

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