一、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]
Constraints:
1 <= nums.length <= 105
-104 <= nums[i] <= 104
1 <= k <= nums.length
二、题解
方法一：优先级队列法
class Solution {
public:
vector<int> maxSlidingWindow(vector<int>& nums, int k) {
int n = nums.size();
priority_queue<pair<int,int>> q;
for(int i = 0;i < k;i++){
q.emplace(nums[i],i);
}
vector<int> res = {
q.top().first};
for(int i = k;i < n;i++){
q.emplace(nums[i],i);
//若最大值不在窗口中，不断移除元素
while(q.top().second <= i-k) q.pop();
res.push_back(q.top().first);
}
return res;
}
};
方法二、单调队列法
class Solution {
public:
vector<int> maxSlidingWindow(vector<int>& nums, int k) {
int n = nums.size();
deque<int> q;
for(int i = 0;i < k;i++){
while(!q.empty() && nums[i] > nums[q.back()]) q.pop_back();
q.push_back(i);
}
vector<int> res = {
nums[q.front()]};
for(int i = k;i < n;i++){
while(!q.empty() && nums[i] > nums[q.back()]) q.pop_back();
q.push_back(i);
while(!q.empty() && q.front() <= i - k) q.pop_front();
res.push_back(nums[q.front()]);
}
return res;
}
};
