Array - 239. Sliding Window Maximum

239. Sliding Window Maximum

Given an array 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:

Input: nums = [1,3,-1,-3,5,3,6,7], and 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

Note You may assume k is always valid, 1 ≤ k ≤ input array's size for non-empty array.

Follow up: Could you solve it in linear time?

思路：

使用monotonic Queue 来做

代码：

java：

class Solution {

    public int[] maxSlidingWindow(int[] nums, int k) {
        if (nums == null || k <= 0 ) return new int[]{};
        
        int len = nums.length;
        int[] res = new int[len - k + 1];
        int index = 0;
        
        Deque<Integer> deque = new ArrayDeque<>();
        for(int i = 0; i < len; i++){
            while(!deque.isEmpty() && deque.peek() < i - k +1) deque.poll();
            while(!deque.isEmpty() && nums[deque.peekLast()] < nums[i]) deque.pollLast();
            
            deque.offer(i);
            if (i >= k-1) {
                res[index++] = nums[deque.peek()];
            }                          
        }
        return res;
    }                                       
}