[LeetCode] 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.

For example,
Given nums = [1,3,-1,-3,5,3,6,7], and k = 3.

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

Therefore, return the max sliding window as [3,3,5,5,6,7].

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

Follow up:
Could you solve it in linear time?

Hint:

1. How about using a data structure such as deque (double-ended queue)?
2. The queue size need not be the same as the window’s size.
3. Remove redundant elements and the queue should store only elements that need to be considered.

解题思路

依据提示，借助一个deque来实现，deque里面存储可能的最大值的下标。遍历数组，若当前元素大于队列尾部元素，则移除尾部元素，然后添加当前元素的下标。提取deque中的值时，需要考虑deque首部的值是否在当前元素的窗口内。

实现代码

// Runtime: 31 ms
public class Solution {
    private Deque<Integer> deque = new LinkedList<Integer>();
    public int[] maxSlidingWindow(int[] nums, int k) {
        if (k == 0) {
            return new int[0];
        }
        int[] res = new int[nums.length - k + 1];
        for (int i = 0; i < nums.length; i++) {
            while (!deque.isEmpty() && nums[i] >= nums[deque.peekLast()]) {
                deque.pollLast();
            }
            deque.offerLast(i);
            while (deque.peekFirst() < i - k + 1) {
                deque.pollFirst();
            }
            if (i >= k - 1) {
                res[i - k + 1] = nums[deque.peekFirst()];
            }
        }

        return res;
    }
}