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| 1 | +/* |
| 2 | + * @lc app=leetcode.cn id=239 lang=cpp |
| 3 | + * |
| 4 | + * [239] 滑动窗口最大值 |
| 5 | + * |
| 6 | + * https://leetcode.cn/problems/sliding-window-maximum/description/ |
| 7 | + * |
| 8 | + * algorithms |
| 9 | + * Hard (49.69%) |
| 10 | + * Likes: 2309 |
| 11 | + * Dislikes: 0 |
| 12 | + * Total Accepted: 443.7K |
| 13 | + * Total Submissions: 892.9K |
| 14 | + * Testcase Example: '[1,3,-1,-3,5,3,6,7]\n3' |
| 15 | + * |
| 16 | + * 给你一个整数数组 |
| 17 | + * nums,有一个大小为 k 的滑动窗口从数组的最左侧移动到数组的最右侧。你只可以看到在滑动窗口内的 |
| 18 | + * k 个数字。滑动窗口每次只向右移动一位。 |
| 19 | + * |
| 20 | + * 返回 滑动窗口中的最大值 。 |
| 21 | + * |
| 22 | + * |
| 23 | + * |
| 24 | + * 示例 1: |
| 25 | + * |
| 26 | + * |
| 27 | + * 输入:nums = [1,3,-1,-3,5,3,6,7], k = 3 |
| 28 | + * 输出:[3,3,5,5,6,7] |
| 29 | + * 解释: |
| 30 | + * 滑动窗口的位置 最大值 |
| 31 | + * --------------- ----- |
| 32 | + * [1 3 -1] -3 5 3 6 7 3 |
| 33 | + * 1 [3 -1 -3] 5 3 6 7 3 |
| 34 | + * 1 3 [-1 -3 5] 3 6 7 5 |
| 35 | + * 1 3 -1 [-3 5 3] 6 7 5 |
| 36 | + * 1 3 -1 -3 [5 3 6] 7 6 |
| 37 | + * 1 3 -1 -3 5 [3 6 7] 7 |
| 38 | + * |
| 39 | + * |
| 40 | + * 示例 2: |
| 41 | + * |
| 42 | + * |
| 43 | + * 输入:nums = [1], k = 1 |
| 44 | + * 输出:[1] |
| 45 | + * |
| 46 | + * |
| 47 | + * |
| 48 | + * |
| 49 | + * 提示: |
| 50 | + * |
| 51 | + * |
| 52 | + * 1 <= nums.length <= 10^5 |
| 53 | + * -10^4 <= nums[i] <= 10^4 |
| 54 | + * 1 <= k <= nums.length |
| 55 | + * |
| 56 | + * |
| 57 | + */ |
| 58 | + |
| 59 | +#include <queue> |
| 60 | +#include <utility> |
| 61 | +#include <vector> |
| 62 | +using namespace std; |
| 63 | + |
| 64 | +// @lc code=start |
| 65 | +class Solution { |
| 66 | +public: |
| 67 | + // 大顶堆,每次遍历先入队,然后判断堆顶是否已经超出滑动窗口 |
| 68 | + vector<int> maxSlidingWindow(vector<int> &nums, int k) { |
| 69 | + vector<int> ret; |
| 70 | + ret.reserve(nums.size() - k); |
| 71 | + priority_queue<pair<int, int>> q; |
| 72 | + for (int i = 0; i < k; i++) { |
| 73 | + q.emplace(nums[i], i); |
| 74 | + } |
| 75 | + ret.push_back(q.top().first); |
| 76 | + for (int i = k; i < nums.size(); i++) { |
| 77 | + q.emplace(nums[i], i); |
| 78 | + while (q.top().second <= i - k) { // 上面已加,所以等于也要丢 |
| 79 | + q.pop(); |
| 80 | + } |
| 81 | + ret.push_back(q.top().first); |
| 82 | + } |
| 83 | + return ret; |
| 84 | + } |
| 85 | +}; |
| 86 | +// @lc code=end |
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