[LeetCode] 239. Sliding Window Maximum

[LeetCode] 239. Sliding Window Maximum

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

思路

对于在同一个窗口的元素 x[i], x[j](x < j)：

1. 若 x[i] < x[j], 则 x[i] 不可能为当前窗口的最大值，同样也不可能是后面窗口的最大值，故 x[i] 可以丢掉。

2. 若 x[j] < x[i], x[j] 可能是后面窗口的最大值（因为 j 位置靠后）。

这样就可以维护一个单调递减的队列。

代码

class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        int n = nums.size();
        vector<int> ans;
        int head = 0, tail = -1;
        int q[n + 1];
        for (int i = 0; i < n; i++) {
            while (tail >= head && i - q[head] + 1 > k) head++;
            while (tail >= head && nums[q[tail]] <= nums[i]) tail--;
            q[++tail] = i;
            if (i >= k-1) ans.push_back(nums[q[head]]);

        }
        return ans;

    }
};