Merge pull request #1743 from wang2jun/master

增加 0583 两个字符串的删除操作 Javascript 方法二

problems/0583.两个字符串的删除操作.md

@@ -228,29 +228,44 @@ func min(a, b int) int {
 ```
 Javascript：
 ```javascript
-const minDistance = (word1, word2) => {
+// 方法一
-    let dp = Array.from(new Array(word1.length + 1), () => Array(word2.length+1).fill(0));
+var minDistance = (word1, word2) => {
-
+  let dp = Array.from(new Array(word1.length + 1), () =>
+    Array(word2.length + 1).fill(0)
+  );
   for (let i = 1; i <= word1.length; i++) {
     dp[i][0] = i;
   }
-
   for (let j = 1; j <= word2.length; j++) {
     dp[0][j] = j;
   }
-
   for (let i = 1; i <= word1.length; i++) {
     for (let j = 1; j <= word2.length; j++) {
       if (word1[i - 1] === word2[j - 1]) {
         dp[i][j] = dp[i - 1][j - 1];
       } else {
-                dp[i][j] = Math.min(dp[i-1][j] + 1, dp[i][j-1] + 1, dp[i-1][j-1] + 2);
+        dp[i][j] = Math.min(
+          dp[i - 1][j] + 1,
+          dp[i][j - 1] + 1,
+          dp[i - 1][j - 1] + 2
+        );
       }
     }
   }
-    
   return dp[word1.length][word2.length];
 };
+
+// 方法二
+var minDistance = function (word1, word2) {
+  let dp = new Array(word1.length + 1)
+    .fill(0)
+    .map((_) => new Array(word2.length + 1).fill(0));
+  for (let i = 1; i <= word1.length; i++)
+    for (let j = 1; j <= word2.length; j++)
+      if (word1[i - 1] === word2[j - 1]) dp[i][j] = dp[i - 1][j - 1] + 1;
+      else dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
+  return word1.length + word2.length - dp[word1.length][word2.length] * 2;
+};
 ```
 
 TypeScript：