From 1516dc3d152b818f1e3d4f3675924c35918e0558 Mon Sep 17 00:00:00 2001
From: resyon <hirosakae7@gmail.com>
Date: Sun, 30 May 2021 21:27:11 +0800
Subject: [PATCH 1/3] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=E4=BA=860239-golang?=
 =?UTF-8?q?=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 problems/0239.滑动窗口最大值.md | 30 ++++++++++++++++++++++++++
 1 file changed, 30 insertions(+)

diff --git a/problems/0239.滑动窗口最大值.md b/problems/0239.滑动窗口最大值.md
index 6cc355ca..ed17157a 100644
--- a/problems/0239.滑动窗口最大值.md
+++ b/problems/0239.滑动窗口最大值.md
@@ -267,6 +267,36 @@ Python：
 
 Go：
 
+```go
+func maxSlidingWindow(nums []int, k int) []int {
+    var queue []int
+    var rtn []int
+
+    for f := 0; f < len(nums); f++ {
+        //维持队列递减, 将 k 插入合适的位置, queue中 <=k 的 元素都不可能是窗口中的最大值, 直接弹出
+        for len(queue) > 0 && nums[f] > nums[queue[len(queue)-1]] {
+            queue = queue[:len(queue)-1]
+        }
+        // 等大的后来者也应入队
+        if len(queue) == 0 || nums[f] <= nums[queue[len(queue)-1]] {
+            queue = append(queue, f)
+        }
+
+        if f >= k - 1 {
+            rtn = append(rtn, nums[queue[0]])
+            //弹出离开窗口的队首
+            if f - k + 1 == queue[0] {
+                queue = queue[1:]
+            }
+        }
+    }
+
+    return rtn
+
+}
+
+```
+
 Javascript:
 ```javascript
 var maxSlidingWindow = function (nums, k) {

From 0de459b1ff5f491e13c4fec93cbc7727dabc73e6 Mon Sep 17 00:00:00 2001
From: LiangDazhu <42199191+LiangDazhu@users.noreply.github.com>
Date: Sun, 30 May 2021 22:54:01 +0800
Subject: [PATCH 2/3] =?UTF-8?q?Update=20=E8=83=8C=E5=8C=85=E7=90=86?=
 =?UTF-8?q?=E8=AE=BA=E5=9F=BA=E7=A1=8001=E8=83=8C=E5=8C=85-2.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

dp[j]可以通过dp[j - weight[i]]推导出来, 应该是写错了吧
---
 problems/背包理论基础01背包-2.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/problems/背包理论基础01背包-2.md b/problems/背包理论基础01背包-2.md
index a169425a..e831d88f 100644
--- a/problems/背包理论基础01背包-2.md
+++ b/problems/背包理论基础01背包-2.md
@@ -55,7 +55,7 @@
 
 dp[j]为 容量为j的背包所背的最大价值，那么如何推导dp[j]呢？
 
-dp[j]可以通过dp[j - weight[j]]推导出来，dp[j - weight[i]]表示容量为j - weight[i]的背包所背的最大价值。
+dp[j]可以通过dp[j - weight[i]]推导出来，dp[j - weight[i]]表示容量为j - weight[i]的背包所背的最大价值。
 
 dp[j - weight[i]] + value[i] 表示 容量为 j - 物品i重量 的背包 加上 物品i的价值。（也就是容量为j的背包，放入物品i了之后的价值即：dp[j]）
 

From 60ad3b081134174577ad58416d275b1bb78267cd Mon Sep 17 00:00:00 2001
From: fusunx <1102654482@qq.com>
Date: Sun, 30 May 2021 23:37:15 +0800
Subject: [PATCH 3/3] =?UTF-8?q?0045.=E8=B7=B3=E8=B7=83=E6=B8=B8=E6=88=8F||?=
 =?UTF-8?q?.md=20Javascript?=
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Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 problems/0045.跳跃游戏II.md | 20 ++++++++++++++++++++
 1 file changed, 20 insertions(+)

diff --git a/problems/0045.跳跃游戏II.md b/problems/0045.跳跃游戏II.md
index 31b52b31..4128da4c 100644
--- a/problems/0045.跳跃游戏II.md
+++ b/problems/0045.跳跃游戏II.md
@@ -208,6 +208,26 @@ func jump(nums []int) int {
 	}
 	return dp[len(nums)-1]
 }
+```
+
+Javascript:
+```Javascript
+var jump = function(nums) {
+    let curIndex = 0
+    let nextIndex = 0
+    let steps = 0
+    for(let i = 0; i < nums.length - 1; i++) {
+        nextIndex = Math.max(nums[i] + i, nextIndex)
+        if(i === curIndex) {
+            curIndex = nextIndex
+            steps++
+        }
+    }
+
+    return steps
+};
+```
+
 /*
 dp[i]表示从起点到当前位置的最小跳跃次数
 dp[i]=min(dp[j]+1,dp[i]) 表示从j位置用一步跳跃到当前位置，这个j位置可能有很多个，却最小一个就可以