LeetCode-Sliding Window Maximum

Posted 2020-11-21 IncredibleThings

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?

大概思路是用双向队列保存数字的下标，遍历整个数组，如果此时队列的首元素是i - k的话，表示此时窗口向右移了一步，则移除队首元素。然后比较队尾元素和将要进来的值，如果小的话就都移除，然后此时我们把队首元素加入结果中即可

class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        if(nums == null || nums.length == 0 || k <= 0 ){
            return new int[0];
        }
        
        int[] res = new int[nums.length-k+1];
        Deque<Integer> deque = new LinkedList<>();
        for(int i=0; i<nums.length; i++){
            
            if(!deque.isEmpty() && deque.getFirst() == i-k){
                deque.removeFirst();
                
            }
            
            while(!deque.isEmpty() && nums[deque.getLast()] <= nums[i]){
                deque.removeLast();
            }
            deque.addLast(i);    
            if(i >= k-1){
                res[i+1-k] = nums[deque.getFirst()];
            }
            
        }
        return res;
    }
}