feat: add python and java solutions to lcof problem: No.59II

添加《剑指 Offer》题解：剑指 Offer 59 - II. 队列的最大值

Author: yanglbme

README.md

@@ -58,18 +58,20 @@
 
 非常感谢以下所有朋友对本项目的贡献，你们是最可爱的人！
 
-<a href="https://opencollective.com/doocs-leetcode/contributors.svg?width=890&button=true"><img src="https://opencollective.com/doocs-leetcode/contributors.svg?width=1200&button=false" /></a>
+<a href="https://opencollective.com/doocs-leetcode/contributors.svg?width=890&button=true" target="_blank"><img src="https://opencollective.com/doocs-leetcode/contributors.svg?width=1200&button=false" /></a>
 
 ## 赞助人
 
-<a href="https://opencollective.com/doocs-leetcode#backers" target="_blank"><img src="https://opencollective.com/doocs-leetcode/backers.svg?width=890"></a>
+<a href="https://opencollective.com/doocs-leetcode/backers.svg?width=890" target="_blank"><img src="https://opencollective.com/doocs-leetcode/backers.svg?width=890"></a>
 
 > You help the developer community practice for interviews, and there is nothing better we could ask for. -- [Alan Yessenbayev](https://opencollective.com/alan-yessenbayev)
 
 ## 赞助商
 
-<a href="https://opencollective.com/doocs-leetcode/sponsor/0/website" target="_blank"><img src="https://opencollective.com/doocs-leetcode/sponsor/0/avatar.svg"></a>
+<a href="https://opencollective.com/doocs-leetcode/sponsors.svg?width=890" target="_blank"><img src="https://opencollective.com/doocs-leetcode/sponsors.svg?width=890"></a>
 
 ## 许可证
 
-<a rel="license" href="https://github.com/doocs/leetcode/blob/master/LICENSE"><img alt="Creative Commons License" style="border-width:0" src="./img/cc-by-sa-88x31.png" /></a><br /><a rel="license" href="http://creativecommons.org/licenses/by-sa/4.0/">知识共享 版权归属-相同方式共享 4.0 国际 公共许可证</a>
+<a rel="license" href="http://creativecommons.org/licenses/by-sa/4.0/">知识共享 版权归属-相同方式共享 4.0 国际 公共许可证</a>
+
+<a rel="license" href="https://github.com/doocs/leetcode/blob/master/LICENSE"><img alt="Creative Commons License" style="border-width:0" src="./img/cc-by-sa-88x31.png" /></a>

README_EN.md

@@ -60,14 +60,20 @@ This project exists thanks to all the people who contribute.
 
 ## Backers
 
+Thank you to all our backers!
+
 <a href="https://opencollective.com/doocs-leetcode#backers" target="_blank"><img src="https://opencollective.com/doocs-leetcode/backers.svg?width=890"></a>
 
 > You help the developer community practice for interviews, and there is nothing better we could ask for. -- [Alan Yessenbayev](https://opencollective.com/alan-yessenbayev)
 
 ## Sponsors
 
-<a href="https://opencollective.com/doocs-leetcode/sponsor/0/website" target="_blank"><img src="https://opencollective.com/doocs-leetcode/sponsor/0/avatar.svg"></a>
+Support this project by becoming a sponsor. Your logo will show up here with a link to your website.
+
+<a href="https://opencollective.com/doocs-leetcode#sponsors" target="_blank"><img src="https://opencollective.com/doocs-leetcode/sponsors.svg?width=890"></a>
 
 ## License
 
-<a rel="license" href="https://github.com/doocs/leetcode/blob/master/LICENSE"><img alt="Creative Commons License" style="border-width:0" src="./img/cc-by-sa-88x31.png" /></a><br />This work is licensed under a <a rel="license" href="http://creativecommons.org/licenses/by-sa/4.0/">Creative Commons Attribution-ShareAlike 4.0 International License</a>.
+This work is licensed under a <a rel="license" href="http://creativecommons.org/licenses/by-sa/4.0/">Creative Commons Attribution-ShareAlike 4.0 International License</a>.
+
+<a rel="license" href="https://github.com/doocs/leetcode/blob/master/LICENSE"><img alt="Creative Commons License" style="border-width:0" src="./img/cc-by-sa-88x31.png" /></a>

lcof/面试题59 - II. 队列的最大值/README.md

@@ -25,23 +25,86 @@
 ```
 
 **限制：**
+
 - `1 <= push_back,pop_front,max_value的总操作数 <= 10000`
 - `1 <= value <= 10^5`
 
 ## 解法
 <!-- 这里可写通用的实现逻辑 -->
 
+利用一个辅助队列按单调顺序存储当前队列的最大值。
+
 
 ### Python3
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 ```python
-
+class MaxQueue:
+
+    def __init__(self):
+        self.p = []
+        self.q = []
+
+    def max_value(self) -> int:
+        return -1 if not self.q else self.q[0]
+
+    def push_back(self, value: int) -> None:
+        while self.q and self.q[-1] < value:
+            self.q.pop(-1)
+        self.q.append(value)
+        self.p.append(value)
+        
+
+    def pop_front(self) -> int:
+        if not self.p: return -1
+        res = self.p.pop(0)
+        if res == self.q[0]:
+            self.q.pop(0)
+        return res
 ```
 
 ### Java
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 ```java
+class MaxQueue {
+
+    private Queue<Integer> p = new ArrayDeque<>();
+    private Deque<Integer> q = new ArrayDeque<>();
 
+    public MaxQueue() {
+
+    }
+    
+    public int max_value() {
+        return q.isEmpty() ? -1 : q.peekFirst();
+    }
+    
+    public void push_back(int value) {
+        while (!q.isEmpty() && q.peekLast() < value) {
+            q.pollLast();
+        }
+        q.addLast(value);
+        p.add(value);
+    }
+    
+    public int pop_front() {
+        if (p.isEmpty()) {
+            return -1;
+        }
+        int res = p.poll();
+        if (res == q.peekFirst()) {
+            q.pollFirst();
+        }
+        return res;
+    }
+}
+
+/**
+ * Your MaxQueue object will be instantiated and called as such:
+ * MaxQueue obj = new MaxQueue();
+ * int param_1 = obj.max_value();
+ * obj.push_back(value);
+ * int param_3 = obj.pop_front();
+ */
 ```
 
 ### JavaScript

lcof/面试题59 - II. 队列的最大值/Solution.java

@@ -0,0 +1,40 @@
+class MaxQueue {
+
+    private Queue<Integer> p = new ArrayDeque<>();
+    private Deque<Integer> q = new ArrayDeque<>();
+
+    public MaxQueue() {
+
+    }
+    
+    public int max_value() {
+        return q.isEmpty() ? -1 : q.peekFirst();
+    }
+    
+    public void push_back(int value) {
+        while (!q.isEmpty() && q.peekLast() < value) {
+            q.pollLast();
+        }
+        q.addLast(value);
+        p.add(value);
+    }
+    
+    public int pop_front() {
+        if (p.isEmpty()) {
+            return -1;
+        }
+        int res = p.poll();
+        if (res == q.peekFirst()) {
+            q.pollFirst();
+        }
+        return res;
+    }
+}
+
+/**
+ * Your MaxQueue object will be instantiated and called as such:
+ * MaxQueue obj = new MaxQueue();
+ * int param_1 = obj.max_value();
+ * obj.push_back(value);
+ * int param_3 = obj.pop_front();
+ */

lcof/面试题59 - II. 队列的最大值/Solution.py

@@ -0,0 +1,22 @@
+class MaxQueue:
+
+    def __init__(self):
+        self.p = []
+        self.q = []
+
+    def max_value(self) -> int:
+        return -1 if not self.q else self.q[0]
+
+    def push_back(self, value: int) -> None:
+        while self.q and self.q[-1] < value:
+            self.q.pop(-1)
+        self.q.append(value)
+        self.p.append(value)
+        
+
+    def pop_front(self) -> int:
+        if not self.p: return -1
+        res = self.p.pop(0)
+        if res == self.q[0]:
+            self.q.pop(0)
+        return res