0239. Sliding Window Maximum (H)

Sliding Window Maximum (H)

題目

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

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題意

給定一個數組和一個定長窗口，將窗口在數組上從左到右滑動，記錄每一步在當前窗口中的最大值。

思路

1. 優先隊列

   維護一個優先隊列，存儲一個數值對(nums[index], index)。遍歷數組，計算當前窗口的左邊界left，將當前數字加入到優先隊列中，查看當前優先隊列中的最大值的下標是否小於left，若是是則說明該最大值不在當前窗口中，出隊，重複操做直到最大值在當前窗口中，並加入結果集。

2. 雙向隊列

   維護一個雙向隊列，存儲下標。遍歷數組，計算當前窗口的左邊界left，若是隊首元素小於left則出隊；接着從隊尾開始，將全部小於當前元素的下標依次出隊，最後將當前下標入隊。這樣能保證每次剩下的隊首元素都是當前窗口中的最大值。

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代碼實現

Java

優先隊列

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;
    }
}

雙向隊列

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;
    }
}