• 0239. Sliding Window Maximum (H)


    Sliding Window Maximum (H)

    题目

    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 <= 10^5
    • -10^4 <= nums[i] <= 10^4
    • 1 <= k <= nums.length

    题意

    给定一个数组和一个定长窗口,将窗口在数组上从左到右滑动,记录每一步在当前窗口中的最大值。

    思路

    1. 优先队列

      维护一个优先队列,存储一个数值对(nums[index], index)。遍历数组,计算当前窗口的左边界left,将当前数字加入到优先队列中,查看当前优先队列中的最大值的下标是否小于left,如果是则说明该最大值不在当前窗口中,出队,重复操作直到最大值在当前窗口中,并加入结果集。

    2. 双向队列

      维护一个双向队列,存储下标。遍历数组,计算当前窗口的左边界left,如果队首元素小于left则出队;接着从队尾开始,将所有小于当前元素的下标依次出队,最后将当前下标入队。这样能保证每次剩下的队首元素都是当前窗口中的最大值。


    代码实现

    Java

    优先队列

    class Solution {
        public int[] maxSlidingWindow(int[] nums, int k) {
            int[] ans = new int[nums.length - k + 1];
            Queue<int[]> q = new PriorityQueue<>((a, b) -> b[0] - a[0]);
            int left = 0;
    
            for (int i = 0; i < nums.length; i++) {
                left = i - k + 1;
                q.offer(new int[]{nums[i], i});
    
                if (left >= 0) {
                    while (q.peek()[1] < left) {
                        q.poll();
                    }
                    ans[left] = q.peek()[0];
                }
            }
    
            return ans;
        }
    }
    

    双向队列

    class Solution {
        public int[] maxSlidingWindow(int[] nums, int k) {
            int[] ans = new int[nums.length - k + 1];
            Deque<Integer> q = new ArrayDeque<>();
    
            for (int i = 0; i < nums.length; i++) {
                int left = i - k + 1;
    
                if (!q.isEmpty() && q.peekFirst() < left) {
                    q.pollFirst();
                }
                while (!q.isEmpty() && nums[i] > nums[q.peekLast()]) {
                    q.pollLast();
                }
                q.offerLast(i);
    
                if (left >= 0) {
                    ans[left] = nums[q.peekFirst()];
                }
            }
    
            return ans;
        }
    }
    
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  • 原文地址:https://www.cnblogs.com/mapoos/p/14054079.html
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