LeetCode One Question of the Day-2.11-1423-Maximum Available Points C++

Title description

Solve the small root heap to solve the TOP K problem

Small root pile solves TOP K ideas:

Find the minimum

First build a large root pile , and then press k numbers
. Each remaining number is first pressed, and then the top element
of the pile is popped. After all the elements are pressed, the top of the pile is the k-th smallest number.

Find the maximum

First create a small root heap , and then press k numbers
. Each remaining number is first pressed, and then the top element
of the heap is popped. After all the elements are pressed, the top of the heap is the k-th largest number
ps: C++ defaults to the big root heap. Here you can remember the wording of the root of the big and small

• Little with greater

•   priority_queue<int, vector<int>, greater<int>> q;
  

• Radish less

•   priority_queue<int, vector<int>, less<int>> q;
  

class KthLargest {
    
    
public:
    //小根堆 堆首取第k大元素
    priority_queue<int, vector<int>, greater<int>> q;
    int k;
    KthLargest(int k, vector<int>& nums) {
    
    
        this -> k = k;
        this -> q = priority_queue<int, vector<int>, greater<int>> (nums.begin(), nums.end());
    }
    
    int add(int val) {
    
    
        q.push(val);
        if(q.size() < k) return NULL;
        while(q.size() > k) {
    
    
            q.pop();
        }
        return q.top();
    }
};

/**
 * Your KthLargest object will be instantiated and called as such:
 * KthLargest* obj = new KthLargest(k, nums);
 * int param_1 = obj->add(val);
 */

Time complexity O(1) Average complexity of add function
Space complexity O(n)