Given an integer array nums, there is a sliding window of size k that moves from the leftmost side of the array to the rightmost side of the array. You can only see the k numbers in the sliding window. The sliding window only moves one position to the right at a time.
Returns the maximum value in the sliding window.
Example 1:
Input: nums = [1,3,-1,-3,5,3,6,7], k = 3
Output: [3,3,5,5,6,7]
Explanation:
The position of the sliding window Maximum
--------------- -----
[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]
analysis:
The main thing is to learn to use priority_queue. Compared with ordinary queues, emplace can be used to ensure that the top element is the maximum value. The code is as follows
class Solution {
public:
vector<int> maxSlidingWindow(vector<int>& nums, int k) {
int n = nums.size();
priority_queue<pair<int, int>> q;
// 前k个构造一下优先队列
for(int i = 0; i < k; i++){
q.emplace(nums[i], i);
// 大顶堆，这个top对应的就是前k个里面最大的，从大到小的一个队列
// cout << q.top().first << " " << q.top().second ;
}
// 构造一个数组专门装每次窗口的最大值
vector<int> resultMax;
resultMax.push_back(q.top().first);
for(int i = k; i < nums.size(); i++){
q.emplace(nums[i], i);
// 为的是，假如后面窗口中的最大值比前面窗口的最大值小，那么堆顶一直都是原来窗口的最大值，需要把这些值给及时的pop掉，同样也能减轻堆的大小，比如[1, -1] k = 1这类情况
while(q.top().second <= i - k){
q.pop();
}
resultMax.push_back(q.top().first);
}
return resultMax;
}
};
