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Array - 239. Sliding Window Maximum

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;
    }                                       
}

标题:Array - 239. Sliding Window Maximum
作者:michael
地址:https://www.bitbo-liuyang.com/articles/2019/06/23/1561303509044.html

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