【leetcode】1278. Palindrome Partitioning III

题目如下：

You are given a string s containing lowercase letters and an integer k. You need to :

• First, change some characters of s to other lowercase English letters.
• Then divide s into k non-empty disjoint substrings such that each substring is palindrome.

Return the minimal number of characters that you need to change to divide the string.

Example 1:

Input: s = "abc", k = 2
Output: 1
Explanation: You can split the string into "ab" and "c", and change 1 character in "ab" to make it palindrome.

Example 2:

Input: s = "aabbc", k = 3
Output: 0
Explanation: You can split the string into "aa", "bb" and "c", all of them are palindrome.

Example 3:

Input: s = "leetcode", k = 8
Output: 0

Constraints:

• 1 <= k <= s.length <= 100.
• s only contains lowercase English letters.

解题思路：记dp[i][j] = v 表示s[0:j]区间内分割成j段，只需要改变v个字符就可以使得每一段的子串都是回文串。要求dp[i][j]的值，关键就是找出j和j-1的分割点。很显然有dp[i][j] = min(dp[m][j-1] + s[m:j]需要改变的字符的个数)，只需要找出最小的分割点m即可。

代码如下：

class Solution(object):
    def palindromePartition(self, s, k):
        """
        :type s: str
        :type k: int
        :rtype: int
        """
        def calc(substr):
            count = 0
            for i in range(len(substr)/2):
                if substr[i] != substr[len(substr)-i - 1]:count += 1
            return count

        dp = [[float(‘inf‘)] * k for _ in s]
        dp[0][0] = 0
        for i in range(len(s)):
            for j in range(k):
                if j == 0:
                    dp[i][j] = calc(s[:i+1])
                    continue
                for m in range(j-1,i):
                    dp[i][j] = min(dp[i][j],dp[m][j-1] + calc(s[m+1:i+1]))
        #print dp
        return dp[-1][-1]

原文地址：https://www.cnblogs.com/seyjs/p/12000533.html