5278. Palindrome Partitioning III
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.
：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/palindrome-partitioning-iii
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这道题，考的时候不会，所以我来分析一下这个第一的美国老哥的代码
人家三分钟就写完了~~~~我题目都看了十分钟。。。
const int N_MAX = 105;
const int INF = 1e9 + 5;
class Solution {
public:
int dp[N_MAX][N_MAX];
int palindromePartition(string s, int k) {
int n = s.size();
for (int i = 0; i <= n; i++)
for (int j = 0; j <= k; j++)
dp[i][j] = INF;
dp[0][0] = 0;
//i是长度到i,j是分成了j段
for (int i = 0; i < n; i++)
for (int j = 0; j < k; j++)
for (int len = 1; i + len <= n; len++) {
int cost = 0;//cost 是需要转换的个数
for (int a = i, b = i + len - 1; a < b; a++, b--)
cost += s[a] != s[b];//把需要换的部分换了
dp[i + len][j + 1] = min(dp[i + len][j + 1], dp[i][j] + cost);//比较是否需要更新
}
return dp[n][k];
}
};
感觉也是暴力，但是中间一直在更新最小值。
最后找到长度N，分成K份的值就行。
