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

@@ -228,28 +228,43 @@ 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), () =>
-    for(let i = 1; i <= word1.length; i++) {
+    Array(word2.length + 1).fill(0)
-        dp[i][0] = i; 
+  );
+  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
+        );
+      }
     }
+  }
+  return dp[word1.length][word2.length];
+};
 
-    for(let j = 1; j <= word2.length; j++) {
+// 方法二
-        dp[0][j] = j;
+var minDistance = function (word1, word2) {
-    }
+  let dp = new Array(word1.length + 1)
-
+    .fill(0)
-    for(let i = 1; i <= word1.length; i++) {
+    .map((_) => new Array(word2.length + 1).fill(0));
-        for(let j = 1; j <= word2.length; j++) {
+  for (let i = 1; i <= word1.length; i++)
-            if(word1[i-1] === word2[j-1]) {
+    for (let j = 1; j <= word2.length; j++)
-                dp[i][j] = dp[i-1][j-1];
+      if (word1[i - 1] === word2[j - 1]) dp[i][j] = dp[i - 1][j - 1] + 1;
-            } else {
+      else dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
-                dp[i][j] = Math.min(dp[i-1][j] + 1, dp[i][j-1] + 1, dp[i-1][j-1] + 2);
+  return word1.length + word2.length - dp[word1.length][word2.length] * 2;
-            }
-        }
-    }
-    
-    return dp[word1.length][word2.length];
 };
 ```