Sliding Window Maximum Leetcode

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?

 

这道题理解了sliding window的本质还是挺好做的，每次减一个再加一个保证priorityqueue里面只有k个值就可以了。

注意：

Java默认为最小堆，不要忘记改为最大的。另外要注意单独处理最后一个值。

public class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        if (nums.length == 0) {
            return new int[0];
        }
        int[] res = new int[nums.length - k + 1];
        PriorityQueue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder());
        for (int i = 0; i < k; i++) {
            pq.add(nums[i]);
        }
        for (int i = k; i < nums.length; i++) {
            res[i - k] = pq.peek();
            pq.remove(nums[i - k]);
            pq.add(nums[i]);
        }
        res[nums.length - k] = pq.peek();
        return res;
    }
}

follow up要求O(n)，虽然听过但还是用普通的queue做了一遍才知道queue为啥不行。。。用deque的主要思路是每次移走k范围外的数，剩下的数按照从大到小排列。添加一个数的时候，如果queue里面的数比要添加的数小，就移出来。这样每次保证头部的就是结果。

public class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        if (nums.length == 0) {
            return new int[0];
        }
        int[] res = new int[nums.length - k + 1];
        Deque<Integer> dq = new LinkedList<>();
        for (int i = 0; i < nums.length; i++) {
            while (!dq.isEmpty() && dq.peek() <= i - k) {
                dq.poll();
            }
            while (!dq.isEmpty() && nums[dq.peekLast()] <= nums[i]) {
                dq.pollLast();
            }
            dq.offer(i);
            if (i - k + 1 >= 0) {
                res[i - k + 1] = nums[dq.peek()];
            }
        }
        return res;
    }
}

另外发现ArrayDeque比linkedlist快。到时候回顾下这个题目好了。