You are given an array of integers nums, there is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position.
Return the max sliding window.
Example 1:
Input: nums = [1,3,-1,-3,5,3,6,7], k = 3
Output: [3,3,5,5,6,7]
Explanation:
Window position Max
--------------- -----
[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]
Example 3:
Input: nums = [1,-1], k = 1
Output: [1,-1]
Example 4:
Input: nums = [9,11], k = 2
Output: [11]
Example 5:
Input: nums = [4,-2], k = 2
Output: [4]
Constraints:
- 1 <= nums.length <= 105
- -104 <= nums[i] <= 104
- 1 <= k <= nums.length
class Solution {
public int[] maxSlidingWindow(int[] nums, int k) {
if(nums == null || nums.length < 2) return nums;
// 双向队列 保存当前窗口最大值的数组位置 保证队列中数组位置的数值按从大到小排序
LinkedList<Integer> queue = new LinkedList();
// 结果数组
int[] result = new int[nums.length-k+1];
// 遍历nums数组
for(int i = 0;i < nums.length;i++){
// 保证从大到小 如果前面数小则需要依次弹出，直至满足要求
while(!queue.isEmpty() && nums[queue.peekLast()] <= nums[i]){
queue.pollLast();
}
// 添加当前值对应的数组下标
queue.addLast(i);
// 判断当前队列中队首的值是否有效
if(queue.peek() <= i-k){
queue.poll();
}
// 当窗口长度为k时 保存当前窗口中最大值
if(i+1 >= k){
result[i+1-k] = nums[queue.peek()];
}
}
return result;
}
}