1143. 最长公共子序列

src/main.rs

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