From 59fdafc7a116ca694b19ba78721bdc416cda76e8 Mon Sep 17 00:00:00 2001
From: Jack <965555169@qq.com>
Date: Mon, 2 Aug 2021 08:58:19 +0800
Subject: [PATCH 1/2] =?UTF-8?q?=E6=B7=BB=E5=8A=A053.=E6=9C=80=E5=A4=A7?=
 =?UTF-8?q?=E5=AD=90=E5=BA=8F=E5=92=8C(=E5=8A=A8=E6=80=81=E8=A7=84?=
 =?UTF-8?q?=E5=88=92)-JavaScript=E7=89=88?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 .../0053.最大子序和（动态规划）.md   | 17 +++++++++++++++++
 1 file changed, 17 insertions(+)

diff --git a/problems/0053.最大子序和（动态规划）.md b/problems/0053.最大子序和（动态规划）.md
index ada378b2..20a8c66c 100644
--- a/problems/0053.最大子序和（动态规划）.md
+++ b/problems/0053.最大子序和（动态规划）.md
@@ -168,6 +168,23 @@ func max(a,b int) int{
 }
 ```
 
+JavaScript：
+
+```javascript
+const maxSubArray = nums => {
+    // 数组长度，dp初始化
+    const [len, dp] = [nums.length, [nums[0]]];
+    // 最大值初始化为dp[0]
+    let max = dp[0];
+    for (let i = 1; i < len; i++) {
+        dp[i] = Math.max(dp[i - 1] + nums[i], nums[i]);
+        // 更新最大值
+        max = Math.max(max, dp[i]);
+    }
+    return max;
+};
+```
+
 
 
 -----------------------

From 9a87b13fc3f87c1b46310d1c8da92b7b8ca8d078 Mon Sep 17 00:00:00 2001
From: Jack <965555169@qq.com>
Date: Mon, 2 Aug 2021 09:16:10 +0800
Subject: [PATCH 2/2] =?UTF-8?q?=E6=B7=BB=E5=8A=A0392.=E5=88=A4=E6=96=AD?=
 =?UTF-8?q?=E5=AD=90=E5=BA=8F=E5=88=97(=E5=8A=A8=E6=80=81=E8=A7=84?=
 =?UTF-8?q?=E5=88=92)-JavaScript=E7=89=88?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 problems/0392.判断子序列.md | 25 ++++++++++++++++++++++++-
 1 file changed, 24 insertions(+), 1 deletion(-)

diff --git a/problems/0392.判断子序列.md b/problems/0392.判断子序列.md
index d97d2684..09ee2c4d 100644
--- a/problems/0392.判断子序列.md
+++ b/problems/0392.判断子序列.md
@@ -180,7 +180,30 @@ class Solution:
         return False
 ```
 
-Go：
+JavaScript：
+
+```javascript
+const isSubsequence = (s, t) => {
+    // s、t的长度
+    const [m, n] = [s.length, t.length];
+    // dp全初始化为0
+    const dp = new Array(m + 1).fill(0).map(x => new Array(n + 1).fill(0));
+    for (let i = 1; i <= m; i++) {
+        for (let j = 1; j <= n; j++) {
+            // 更新dp[i][j]，两种情况
+            if (s[i - 1] === t[j - 1]) {
+                dp[i][j] = dp[i - 1][j - 1] + 1;
+            } else {
+                dp[i][j] = dp[i][j - 1];
+            }
+        }
+    }
+    // 遍历结束，判断dp右下角的数是否等于s的长度
+    return dp[m][n] === m ? true : false;
+};
+```
+
+