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C + + TOPK problem

2020-12-06 17:44:58 ShellCollector

 

https://blog.csdn.net/SmartDemo/article/details/107572238

https://blog.csdn.net/wuqingshan2010/article/details/108508676

	int K = 5;
	std::vector<float> scores;

	scores.push_back(0.56);//  Press in elements 
	scores.push_back(10.56);//  Press in elements 
	scores.push_back(01.56);//  Press in elements 
	scores.push_back(02.56);//  Press in elements 
	scores.push_back(05.56);//  Press in elements 
	scores.push_back(50.56);//  Press in elements 
	scores.push_back(30.56);//  Press in elements 
	std::vector<size_t> idx(scores.size());
	//std::iota(idx.begin(), idx.end(), 0);
	std::sort(idx.begin(), idx.end(),
		[&scores](size_t index_1, size_t index_2) { return scores[index_1] > scores[index_2]; });
	//  obtain K value 
	int k_num = std::min<int>(scores.size(), K);
	std::vector<float> scores_K;
	int idx_j = 0;
	for (int j = 0; j < k_num; ++j) {
		idx_j = idx[j];
		scores_K.push_back(scores[idx_j]);
	}

Cannot return index :

https://blog.csdn.net/doctor_xiong/article/details/81083942

for example :arr[] = {9,1,6,2,3,8,3,4,7,0} The biggest four elements are 6,7,8,9

Ideas : Using small piles , First of all, the array of K Elements inserted into the heap , And then from the K Start traversing the array , If the elements in the array are greater than , Heap top element , It's going to be the top element pop, And then the elements in the array push Get in the pile

Implementation code :

#include<iostream>
using namespace std;
#include<queue>
class compare{
    public:
    bool operator()(int a,int b){
        return a > b;
    }
};

int* FindTopK(int* arr,int n,int* ret,int m){
    compare com;
    priority_queue<int,vector<int>,greater<int> > q(arr,arr+m);
    int i = m;
    for(;i<n;i++){
        if(arr[i] > q.top()){
            q.pop();
            q.push(arr[i]);
        }
    }
    int j = 0;
    while(!q.empty()){
        ret[j++] = q.top();
        q.pop();
    }
}

int main(){
    int arr[] = {9,4,5,2,5,1,7,3,1,8};
    int ret[5];
    FindTopK(arr,10,ret,5);
    for(int i = 0;i<5;i++)
        cout<<ret[i]<<" ";
    return 0;
}

Reference resources 1:

https://blog.csdn.net/qq_37891889/article/details/88621591

Reference resources 2:

https://blog.csdn.net/weixin_43860854/article/details/108616568

Reference resources 3:

https://blog.csdn.net/propro1314/article/details/43091387

Reference resources :https://blog.csdn.net/Hairy_Monsters/article/details/79776744 

/* Case one —— Binary heap C++ Code implementation */
/*
* The code uses STL The implementation of minimum priority queue in , because STL With minimum priority queue in , The bottom layer is a binary heap implementation ,
* So I don't write binary pile anymore . The top element of the minimum priority queue is always the smallest element in the queue , It's the top of the bifurcated reactor .
*/
 
#include <iostream>
#include <queue>
#include <vector>
using namespace std;
 
/* because STL The built-in priority queue is the default maximum priority , So I wrote a comparison function , Change it to minimum priority */
struct cmp1 {
	bool operator ()(int &a, int &b) {
		return a>b;											// Minimum priority 
	}
};
 
int main() {
	// This is used to test , Input format : First enter the maximum required K Of the number of K value , Then input the data stream in turn 
	int K = 0;
	cin >> K;
	int tmp = 0;
	int i = 0;
	priority_queue<int,vector<int>,cmp1> minHeap;			// Set up the minimum priority queue 
	while (cin >> tmp) {									// Loop in the data stream 
		if (i < K) {										// Create a K Priority queue size , That is to say K The size of the binary pile 
			minHeap.push(tmp);
		}
		else {												// Algorithm implementation 
			if (tmp <= minHeap.top())
				continue;
			else if (tmp > minHeap.top()) {
				minHeap.pop();
				minHeap.push(tmp);
			}
		}
		i++;
	}
	while (!minHeap.empty()) {								// The biggest output K Number 
		cout << minHeap.top() << endl;
		minHeap.pop();
	}
	return 0;
}
 
/* The second case ——Quick Select C++ Code implementation */

/*Quick Select*/
#include <iostream>
#include <vector>
 
using namespace std;
 
int Partition(vector<int> &vec, int p, int r) {				// To achieve fast scheduling Partition function , Enter the original array reference , And the left and right subscripts that need to be run 
	if (p >= r)												// illegal input ,Partition Specific ideas refer to the quick arrangement 
		return r;
	int tmp = vec[r];
	int i = p;
	int j = p;
	while (i < r) {
		if (vec[i] <= tmp) {
			int temp = vec[i];
			vec[i] = vec[j];
			vec[j] = temp;
			i++;
			j++;
		}
		else if (vec[i] > tmp) {
			i++;
		}
	}
	vec[r] = vec[j];
	vec[j] = tmp;
	return j;
}
 
int main() {
	int K = 0;										// Test part , Enter the required K Value size , Then input the array elements in turn 
	cin >> K;
	int tmp = 0;
	vector<int> vec;
	while (cin >> tmp)
		vec.push_back(tmp);
	int size = vec.size();
        if (size == 0 || k>size) return vector<int>();
        if (size== k) return input;
        int p = 0;
	int r = vec.size() - 1;
	int index = Partition(vec, p, r);
	while (index != size - K) {						// When Partition The return value and the right part are not K Big hour , Continue to cycle 
		int sizeOfRight = size - index - 1;			// Record index Right array length size 
		if (K <= sizeOfRight) {
			index = Partition(vec, index + 1, r);
		}
		else if (K == sizeOfRight + 1)				// This step seems a bit redundant ,while The loop guarantees that , But in order to correspond to the blog text description, I added 
			continue;
		else if (K > sizeOfRight + 1) {
			index = Partition(vec, p, index - 1);
		}
	}
	for (int i = index; i < size; i++) {			// Test part , Output what you need K Number 
		cout << vec[i] << endl;
	}
	return 0;
}

 

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