// 队列que是STL的双向队列deque
// 队列存放的是元素在序列中的序号
//参考https://www.jianshu.com/p/e59d51e1eef5
class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        deque<int> deque_1;
        vector<int> result;
 
        for(int i=0; i<nums.size(); i++){
            while(!deque_1.empty() && nums[deque_1.back()] < nums[i]){
                deque_1.pop_back();
            }
            deque_1.push_back(i);
            if(i>=k-1){
                while(!deque_1.empty() && deque_1.front()<i-k+1){
                    deque_1.pop_front();
                }
                result.push_back(nums[deque_1.front()]);
            }
        }
        return result;
    }
};

347. 前 K 个高频元素

class Solution {
public:
    // 小顶堆
    class mycomparison {
    public:
        bool operator()(const pair<int, int>& lhs, const pair<int, int>& rhs) {
            return lhs.second > rhs.second;
        }
    };
 
    vector<int> topKFrequent(vector<int>& nums, int k) {
        unordered_map<int, int> map_1;
        for(int i=0; i<nums.size(); i++)
            map_1[nums[i]]++;
         
        // 对频率排序
        // 定义一个小顶堆，大小为k
        priority_queue<pair<int, int>, vector<pair<int, int>>, mycomparison> pri_que;
 
        // 用固定大小为k的小顶堆，扫面所有频率的数值
        for(unordered_map<int,int>::iterator it=map_1.begin(); it != map_1.end(); it++){
            pri_que.push(*it);
            if (pri_que.size() > k) { // 如果堆的大小大于了K，则队列弹出，保证堆的大小一直为k
                pri_que.pop();
            }
        }
 
        // 找出前K个高频元素，因为小顶堆先弹出的是最小的，所以倒序来输出到数组
        vector<int> result(k);
        for (int i = k - 1; i >= 0; i--) {
            result[i] = pri_que.top().first;
            pri_que.pop();
        }
        return result;
    }
};