1143. 最长公共子序列
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+12
-17
@@ -1,29 +1,24 @@
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mod arr;
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mod arr;
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struct Solution {}
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struct Solution {}
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impl Solution {
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impl Solution {
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pub fn min_distance(word1: String, word2: String) -> i32 {
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pub fn longest_common_subsequence(text1: String, text2: String) -> i32 {
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let mut dp = vec![vec![0; word2.len() + 1]; word1.len() + 1];
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let mut dp = vec![vec![0; text2.len() + 1]; text1.len() + 1];
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let word1 = word1.chars().collect::<Vec<char>>();
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let text1 = text1.chars().collect::<Vec<char>>();
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let word2 = word2.chars().collect::<Vec<char>>();
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let text2 = text2.chars().collect::<Vec<char>>();
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for i in 1..=word1.len() {
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for i in 0..text1.len() {
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dp[i][0] = dp[i - 1][0] + 1;
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for j in 0..text2.len() {
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}
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dp[i + 1][j + 1] = dp[i + 1][j].max(dp[i][j + 1]);
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for i in 1..=word2.len() {
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if text1[i] == text2[j] {
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dp[0][i] = dp[0][i - 1] + 1;
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dp[i + 1][j + 1] = dp[i + 1][j + 1].max(dp[i][j] + 1);
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}
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}
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for i in 0..word1.len() {
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for j in 0..word2.len() {
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dp[i + 1][j + 1] = (dp[i][j] + if word1[i] != word2[j] { 1 } else { 0 })
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.min(dp[i][j + 1] + 1)
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.min(dp[i + 1][j] + 1);
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}
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}
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}
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}
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dp[word1.len()][word2.len()]
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dp[text1.len()][text2.len()]
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}
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}
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}
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}
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fn main() {
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fn main() {
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println!(
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println!(
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"{:?}",
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"{:?}",
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Solution::min_distance("intention".to_string(), "execution".to_string())
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Solution::longest_common_subsequence("abcde".to_string(), "ace".to_string())
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);
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);
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}
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}
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