[Lintcode]108. Palindrome Partitioning II /[Leetcode]132. Palindrome Partitioning II

108. Palindrome Partitioning II / 132. Palindrome Partitioning II

• 本题难度: Medium/Hard

• Topic: DP

  Description

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

Return the minimum cuts needed for a palindrome partitioning of s.

Example:

Input: "aab"
Output: 1
Explanation: The palindrome partitioning ["aa","b"] could be produced using 1 cut.

我的代码

1. 使用DFS超时

import collections
class Solution:
    """
    @param s: A string
    @return: An integer
    """
    #min BFS
    def minCut(self, s):
        # write your code here
        f = 0
        q = collections.deque([[s,f]])
        while q:
            [tmps, f] = q.popleft()
            if self.isValid(tmps):
                return f
            else:
                for i in range(len(tmps),0,-1):
                    if self.isValid(tmps[:i]):
                        q.append([tmps[i:],f+1])
        return 0
            
        
    def isValid(self, s):
        return s==s[::-1]
        # l = len(s)
        # for i in range(l//2):
        #     if s[i]!=s[l-1-i]:
        #         return False
        # return True

别人的代码

DP

    def minCut(self, s):
        cut = [x for x in range(-1,len(s))]
        for i in range(0,len(s)):
            for j in range(i,len(s)):
                if s[i:j] == s[j:i:-1]:
                    cut[j+1] = min(cut[j+1],cut[i]+1)
        return cut[-1]

思路
我的思路：
最短->bfs

DP思路：
某一点，最短距离等于前面的+最后一个到他的回文。