leetcode python palindrome-partitioning-ii

palindrome-partitioning-ii

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”,
Return1since the palindrome partitioning[“aa”,“b”]could be produced using 1 cut.

题意

即形成回文串的最小割数

思路

没要求列出所有种类，只需求一个割数，首选dp。 一切尽在公式dp[i] = min(dp[j] + 1, dp[i])中，这题还要特别注意初始化dp的技巧。

python实现

class Solution:
    def minCut(self, s):
        n = len(s)
        # 最坏打算
        dp = [i-1 for i in range(n+1)]
        # 保证要取到整个字符串，不要被下标所困扰
        for i in range(1, n+1):
            for j in range(i+1):
                tmp = s[j:i]
                # print(tmp)
                if self.isPalindrome(tmp):
                    dp[i] = min(dp[j] + 1, dp[i])
        return dp[n]


    def isPalindrome(self, s):
        if s == s[::-1]:
            return True
        return False


if __name__ == '__main__':
    s = 'aabcc'
    # s = 'aabaa'
    so = Solution()
    print(so.minCut(s))