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Commit a150c61

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fix(0239): 新增一种java写法
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‎problems/0239.滑动窗口最大值.md

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Java:
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```Java
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//解法一
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//自定义数组
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class MyQueue {
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Deque<Integer> deque = new LinkedList<>();
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return res;
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}
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}
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//解法二
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//利用双端队列手动实现单调队列
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/**
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* 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
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* 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)
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*/
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class Solution {
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public int[] maxSlidingWindow(int[] nums, int k) {
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ArrayDeque<Integer> deque = new ArrayDeque<>();
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int n = nums.length;
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int[] res = new int[n - k + 1];
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int idx = 0;
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for(int i = 0; i < n; i++) {
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// 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
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// 1.队列头结点需要在[i - k + 1, k]范围内,不符合则要弹出
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while(!deque.isEmpty() && deque.peek() < i - k + 1){
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deque.poll();
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}
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// 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
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while(!deque.isEmpty() && nums[deque.peekLast()] < nums[i]) {
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deque.pollLast();
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}
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deque.offer(i);
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// 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
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if(i >= k - 1){
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res[idx++] = nums[deque.peek()];
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}
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}
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return res;
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}
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}
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```
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Python:

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