feat: add solution to lc problem: No.1143

No.1143.Longest Common Subsequence

Author: yanglbme

solution/1100-1199/1143.Longest Common Subsequence/README.md

@@ -55,7 +55,7 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
-动态规划法。
+**方法一：动态规划**
 
 定义 `dp[i][j]` 表示 `text1[0:i-1]` 和 `text2[0:j-1]` 的最长公共子序列（闭区间）。
 
@@ -156,6 +156,31 @@ func max(a, b int) int {
 }
 ```
 
+### **JavaScript**
+
+```js
+/**
+ * @param {string} text1
+ * @param {string} text2
+ * @return {number}
+ */
+var longestCommonSubsequence = function (text1, text2) {
+    const m = text1.length;
+    const n = text2.length;
+    const dp = new Array(m + 1).fill(0).map(() => new Array(n + 1).fill(0));
+    for (let i = 1; i <= m; ++i) {
+        for (let j = 1; j <= n; ++j) {
+            if (text1[i - 1] == text2[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 dp[m][n];
+};
+```
+
 ### **...**
 
 ```

solution/1100-1199/1143.Longest Common Subsequence/README_EN.md

@@ -142,6 +142,31 @@ func max(a, b int) int {
 }
 ```
 
+### **JavaScript**
+
+```js
+/**
+ * @param {string} text1
+ * @param {string} text2
+ * @return {number}
+ */
+var longestCommonSubsequence = function (text1, text2) {
+    const m = text1.length;
+    const n = text2.length;
+    const dp = new Array(m + 1).fill(0).map(() => new Array(n + 1).fill(0));
+    for (let i = 1; i <= m; ++i) {
+        for (let j = 1; j <= n; ++j) {
+            if (text1[i - 1] == text2[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 dp[m][n];
+};
+```
+
 ### **...**
 
 ```

solution/1100-1199/1143.Longest Common Subsequence/Solution.js

@@ -0,0 +1,20 @@
+/**
+ * @param {string} text1
+ * @param {string} text2
+ * @return {number}
+ */
+var longestCommonSubsequence = function (text1, text2) {
+    const m = text1.length;
+    const n = text2.length;
+    const dp = new Array(m + 1).fill(0).map(() => new Array(n + 1).fill(0));
+    for (let i = 1; i <= m; ++i) {
+        for (let j = 1; j <= n; ++j) {
+            if (text1[i - 1] == text2[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 dp[m][n];
+};