[Swift]LeetCode1278. 分割回文串 III | Palindrome Partitioning III

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号:山青咏芝(let_us_code)
➤博主域名:https://www.zengqiang.org
➤GitHub地址:https://github.com/strengthen/LeetCode
➤原文地址:https://www.cnblogs.com/strengthen/p/12151551.html
➤如果链接不是山青咏芝的博客园地址,则可能是爬取作者的文章。
➤原文已修改更新!强烈建议点击原文地址阅读!支持作者!支持原创!
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

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.


给你一个由小写字母组成的字符串 s,和一个整数 k。

请你按下面的要求分割字符串:

首先,你可以将 s 中的部分字符修改为其他的小写英文字母。
接着,你需要把 s 分割成 k 个非空且不相交的子串,并且每个子串都是回文串。
请返回以这种方式分割字符串所需修改的最少字符数。

示例 1:

输入:s = "abc", k = 2
输出:1
解释:你可以把字符串分割成 "ab" 和 "c",并修改 "ab" 中的 1 个字符,将它变成回文串。
示例 2:

输入:s = "aabbc", k = 3
输出:0
解释:你可以把字符串分割成 "aa"、"bb" 和 "c",它们都是回文串。
示例 3:

输入:s = "leetcode", k = 8
输出:0
 

提示:

1 <= k <= s.length <= 100

原文地址:https://www.cnblogs.com/strengthen/p/12151551.html