61 lines
1.6 KiB
Rust
61 lines
1.6 KiB
Rust
/**
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* [1278] Palindrome Partitioning III
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*/
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pub struct Solution {}
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// submission codes start here
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impl Solution {
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pub fn palindrome_partition(s: String, k: i32) -> i32 {
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let s: Vec<char> = s.chars().collect();
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let length = s.len();
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// cost[i][j]
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// 将 s[i..=j] 修改为回文串需要的修改次数
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let mut cost = vec![vec![0; length]; length];
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for span in 2..=length {
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for i in 0..=length - span {
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let j = i + span - 1;
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cost[i][j] = cost[i + 1][j - 1] + if s[i] == s[j] { 0 } else { 1 };
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}
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}
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let k = k as usize;
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// dp[i][j]
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// 将s[..i]分割为j个回文串需要的修改次数
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let mut dp = vec![vec![i32::MAX; k + 1]; length + 1];
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for i in 1..=length {
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for j in 1..=i.min(k) {
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if j == 1 {
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dp[i][j] = cost[0][i - 1];
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} else {
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// 这里k从j - 1开始枚举是因为前面需要进行j - 1次分割时
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// 字符串的长度必须大于等于j - 1
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for k in j - 1..i {
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dp[i][j] = dp[i][j].min(dp[k][j - 1] + cost[k][i - 1]);
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}
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}
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}
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}
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dp[length][k]
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}
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}
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// submission codes end
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#[cfg(test)]
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mod tests {
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use super::*;
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#[test]
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fn test_1278() {
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assert_eq!(1, Solution::palindrome_partition("abc".to_owned(), 2));
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assert_eq!(0, Solution::palindrome_partition("aabbc".to_owned(), 3));
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assert_eq!(0, Solution::palindrome_partition("leetcode".to_owned(), 8));
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}
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}
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