0239. Sliding Window Maximum

https://leetcode.com/problems/sliding-window-maximum

Description

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

ac

class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        // edge cases
        if (nums == null || nums.length == 0 || k < 1 || k > nums.length)
            return new int[0];
        int n = nums.length;

        // process
        int[] res = new int[n - k + 1];
        int resIdx = 0;
        Deque<Integer> dq = new LinkedList<Integer>();
        for (int i  = 0; i < n; i++) {

            // delete head when out of range
            if (!dq.isEmpty() && dq.peek() < i - k + 1) {
                dq.poll();
            }

            // if nums[i] > element in dq, delete element.
            while (!dq.isEmpty() && nums[dq.peekLast()] <= nums[i]) {
                dq.pollLast();
            }
            dq.offer(i);

            // is a window
            if (i >= k - 1) {
                res[resIdx++] = nums[dq.peek()];
            }
        }

        return res;
    }
}