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1. Topic description
You are given an array of integers numswith a k sliding window of size moving from the far left of the array to the far right of the array. k You can only see the numbers in the sliding window . The sliding window moves to the right only one bit at a time.
Returns the maximum value in the sliding window .
Example 1:
Input: nums = [1,3,-1,-3,5,3,6,7], k = 3 Output: [3,3,5,5,6,7] Explanation: The maximum position of the sliding window --------------- ----- [1 3 -1] -3 5 3 6 7 3 1 [3 -1 -3] 5 3 6 7 3 1 3 [-1 -3 5] 3 6 7 5 1 3 -1 [-3 5 3] 6 7 5 1 3 -1 -3 [5 3 6] 7 6 1 3 -1 -3 5 [3 6 7] 7
Example 2:
Input: nums = [1], k = 1 Output: [1]
2. Problem-solving ideas
1. This question is a typical question of sliding windows, but the conventional thinking will make the time complexity relatively large and timeout , so the two-way queue ArrayDeque is used to optimize the algorithm.
2. First, the window is regarded as a container with a capacity of k, and the left side is the head, and it must be the largest, otherwise the queue will be emptied gradually to ensure that the head of the queue is the largest .
3. Put the array elements into the container first, and the first element can only be the position in the array that is just at the bottom of the container , such as
k=3,num={3,4,7,9,4,8}, only 7 can be used as the head element.
4. Finally, save and return the maximum value in the qualified window.
5. What is recorded is the subscript in the array .
3. Realize the code
public int[] maxSlidingWindow(int[] nums, int k) {//滑动窗口最大值 ArrayDeque <Integer>queue=new ArrayDeque<>(); int L=nums.length; int index=0;//数组下标 if (L==1) return nums; int res[]=new int[L-k+1];//结果个数为L-k+1个 for (int i=0;i<L;i++) { while (!queue.isEmpty()&&queue.peek()<i-k+1)//即使已经进入队列但是不能作为队头即移出 queue.poll(); while (!queue.isEmpty()&&nums[queue.peekLast()]<nums[i])//后续元素有大于队头即把容器全部清空 queue.pollLast(); queue.offer(i);//逐步放入容器 if (i>=k-1) { res[index++]=nums[queue.peek()];//记录每次窗口移动最大的元素 } } return res; }
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