leetcode @python 132. Palindrome Partitioning II

题目链接

https://leetcode.com/problems/palindrome-partitioning-ii/

题目原文

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.

For example, given s = "aab",
Return 1 since the palindrome partitioning ["aa","b"] could be produced using 1 cut.

题目大意

承接上一题的题意，求可以满足回文子串切割的最少割数

解题思路

使用动态规划进行求解：用数组dp[i]记录从第0位到i位最小割数，使用i-1对第i个位置进行初始化，如果子串s[j:i]是回文串，则dp[i] = min(dp[i],dp[j]+1)

代码

class Solution(object):
    def minCut(self, s):
        """
        :type s: str
        :rtype: int
        """
        n = len(s)
        dp = [(i - 1) for i in range(n + 1)]
        for i in range(1, n + 1):
            for j in range(i):
                tmp = s[j:i]
                if tmp == tmp[::-1]:
                    dp[i] = min(dp[i], dp[j] + 1)
        return dp[n]