From ac40cc8be29fdbf22b2b6a5b889315de74f602c3 Mon Sep 17 00:00:00 2001
From: KailokFung <jialuo_feng@foxmail.com>
Date: Wed, 23 Jun 2021 10:24:40 +0800
Subject: [PATCH 1/3] =?UTF-8?q?fix(1047):=20=E6=96=B0=E5=A2=9E=E5=8F=8C?=
 =?UTF-8?q?=E6=8C=87=E9=92=88java=E5=86=99=E6=B3=95?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 ...除字符串中的所有相邻重复项.md | 23 +++++++++++++++++++
 1 file changed, 23 insertions(+)

diff --git a/problems/1047.删除字符串中的所有相邻重复项.md b/problems/1047.删除字符串中的所有相邻重复项.md
index 305a287d..4f13cda3 100644
--- a/problems/1047.删除字符串中的所有相邻重复项.md
+++ b/problems/1047.删除字符串中的所有相邻重复项.md
@@ -171,6 +171,29 @@ class Solution {
 }
 ```
 
+拓展：双指针
+```java
+class Solution {
+    public String removeDuplicates(String s) {
+        char[] ch = s.toCharArray();
+        int fast = 0;
+        int slow = 0;
+        while(fast < s.length()){
+            // 直接用fast指针覆盖slow指针的值
+            ch[slow] = ch[fast];
+            // 遇到前后相同值的，就跳过，即slow指针后退一步，下次循环就可以直接被覆盖掉了
+            if(slow > 0 && ch[slow] == ch[slow - 1]){
+                slow--;
+            }else{
+                slow++;
+            }
+            fast++;
+        }
+        return new String(ch,0,slow);
+    }
+}
+```
+
 Python：
 ```python3
 class Solution:

From 625558094da614c5f9a2b358f3291bf102f7e1a7 Mon Sep 17 00:00:00 2001
From: KailokFung <jialuo_feng@foxmail.com>
Date: Wed, 23 Jun 2021 10:31:45 +0800
Subject: [PATCH 2/3] =?UTF-8?q?fix(1047):=20=E4=BD=BF=E7=94=A8ArrayDeque?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 problems/1047.删除字符串中的所有相邻重复项.md | 4 +++-
 1 file changed, 3 insertions(+), 1 deletion(-)

diff --git a/problems/1047.删除字符串中的所有相邻重复项.md b/problems/1047.删除字符串中的所有相邻重复项.md
index 4f13cda3..760a0946 100644
--- a/problems/1047.删除字符串中的所有相邻重复项.md
+++ b/problems/1047.删除字符串中的所有相邻重复项.md
@@ -127,7 +127,9 @@ Java：
 ```Java
 class Solution {
     public String removeDuplicates(String S) {
-        Deque<Character> deque = new LinkedList<>();
+        //ArrayDeque会比LinkedList在除了删除元素这一点外会快一点
+        //参考：https://stackoverflow.com/questions/6163166/why-is-arraydeque-better-than-linkedlist
+        ArrayDeque<Character> deque = new ArrayDeque<>();
         char ch;
         for (int i = 0; i < S.length(); i++) {
             ch = S.charAt(i);

From a150c61f9db97521b2b758affe64e202d6a821d4 Mon Sep 17 00:00:00 2001
From: KailokFung <jialuo_feng@foxmail.com>
Date: Wed, 23 Jun 2021 18:01:20 +0800
Subject: [PATCH 3/3] =?UTF-8?q?fix(0239):=20=E6=96=B0=E5=A2=9E=E4=B8=80?=
 =?UTF-8?q?=E7=A7=8Djava=E5=86=99=E6=B3=95?=
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 | 35 ++++++++++++++++++++++++++
 1 file changed, 35 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 d108335b..ad54a940 100644
--- a/problems/0239.滑动窗口最大值.md
+++ b/problems/0239.滑动窗口最大值.md
@@ -208,6 +208,7 @@ public:
 
 Java：
 ```Java
+//解法一
 //自定义数组
 class MyQueue {
     Deque<Integer> deque = new LinkedList<>();
@@ -260,6 +261,40 @@ class Solution {
         return res;
     }
 }
+
+//解法二
+//利用双端队列手动实现单调队列
+/**
+ * 用一个单调队列来存储对应的下标，每当窗口滑动的时候，直接取队列的头部指针对应的值放入结果集即可
+ * 单调队列类似 （tail -->） 3 --> 2 --> 1 --> 0 (--> head) (右边为头结点，元素存的是下标)
+ */
+class Solution {
+    public int[] maxSlidingWindow(int[] nums, int k) {
+        ArrayDeque<Integer> deque = new ArrayDeque<>();
+        int n = nums.length;
+        int[] res = new int[n - k + 1];
+        int idx = 0;
+        for(int i = 0; i < n; i++) {
+            // 根据题意，i为nums下标，是要在[i - k + 1, k] 中选到最大值，只需要保证两点
+            // 1.队列头结点需要在[i - k + 1, k]范围内，不符合则要弹出
+            while(!deque.isEmpty() && deque.peek() < i - k + 1){
+                deque.poll();
+            }
+            // 2.既然是单调，就要保证每次放进去的数字要比末尾的都大，否则也弹出
+            while(!deque.isEmpty() && nums[deque.peekLast()] < nums[i]) {
+                deque.pollLast();
+            }
+
+            deque.offer(i);
+
+            // 因为单调，当i增长到符合第一个k范围的时候，每滑动一步都将队列头节点放入结果就行了
+            if(i >= k - 1){
+                res[idx++] = nums[deque.peek()];
+            }
+        }
+        return res;
+    }
+}
 ```
 
 Python：