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


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    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

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