使用优先级队列的未排序数组中的第 K 个最小元素
原文:https://www . geesforgeks . org/k-th-未排序数组中的最小元素-使用优先级-队列/
给定一个由 N 个整数和一个整数 K 组成的数组 arr[] ，任务是使用优先级队列找到数组中的 K 第 个最小元素。
示例:
输入: arr[] = {5，20，10，7，1}，N = 5，K = 2 输出: 5 说明:在给定数组中，2 nd 最小元素为 5。因此，所需的输出为 5。
输入: arr[] = {5，20，10，7，1}，N = 5，K = 5 输出: 20 说明:在给定数组中，第 5 个最小元素为 20。因此，所需的输出为 20。
方法:思路是使用 Java 中的priority queue Collection或 priority_queue STL 库实现 Max_Heap 找到 K th 最小数组元素。按照以下步骤解决问题:
- 使用优先级队列实现最大堆。
- 先将 K 阵元推入优先级 _ 队列 。
- 从那时起，每次插入数组元素后，将该元素放在优先级队列的顶部。
- 完成数组的遍历后，将优先队列顶部的元素打印为所需答案。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find kth smallest array element
void kthSmallest(vector<int>& v, int N, int K)
{
// Implement Max Heap using
// a Priority Queue
priority_queue<int> heap1;
for (int i = 0; i < N; ++i) {
// Insert elements into
// the priority queue
heap1.push(v[i]);
// If size of the priority
// queue exceeds k
if (heap1.size() > K) {
heap1.pop();
}
}
// Print the k-th smallest element
cout << heap1.top() << endl;
}
// Driver code
int main()
{
// Given array
vector<int> vec = { 5, 20, 10, 7, 1 };
// Size of array
int N = vec.size();
// Given K
int K = 2;
// Function Call
kthSmallest(vec, N, K % N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.util.*;
class CustomComparator implements Comparator<Integer> {
@Override
public int compare(Integer number1, Integer number2) {
int value = number1.compareTo(number2);
// elements are sorted in reverse order
if (value > 0) {
return -1;
}
else if (value < 0) {
return 1;
}
else {
return 0;
}
}
}
class GFG{
// Function to find kth smallest array element
static void kthSmallest(int []v, int N, int K)
{
// Implement Max Heap using
// a Priority Queue
PriorityQueue<Integer> heap1 = new PriorityQueue<Integer>(new CustomComparator());
for (int i = 0; i < N; ++i) {
// Insert elements into
// the priority queue
heap1.add(v[i]);
// If size of the priority
// queue exceeds k
if (heap1.size() > K) {
heap1.remove();
}
}
// Print the k-th smallest element
System.out.print(heap1.peek() +"\n");
}
// Driver code
public static void main(String[] args)
{
// Given array
int []vec = { 5, 20, 10, 7, 1 };
// Size of array
int N = vec.length;
// Given K
int K = 2;
// Function Call
kthSmallest(vec, N, K % N);
}
}
// This code is contributed by Amit Katiyar
Python 3
# Python3 program for the above approach
# Function to find kth smallest array element
def kthSmallest(v, N, K):
# Implement Max Heap using
# a Priority Queue
heap1 = []
for i in range(N):
# Insert elements into
# the priority queue
heap1.append(v[i])
# If size of the priority
# queue exceeds k
if (len(heap1) > K):
heap1.sort()
heap1.reverse()
del heap1[0]
# Print the k-th smallest element
heap1.sort()
heap1.reverse()
print(heap1[0])
# Driver code
# Given array
vec = [ 5, 20, 10, 7, 1 ]
# Size of array
N = len(vec)
# Given K
K = 2
# Function Call
kthSmallest(vec, N, K % N)
# This code is contributed by divyeshrabadiya07
C
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to find kth smallest array element
static void kthSmallest(int []v, int N, int K)
{
// Implement Max Heap using
// a Priority Queue
List<int> heap1 = new List<int>();
for (int i = 0; i < N; ++i) {
// Insert elements into
// the priority queue
heap1.Add(v[i]);
// If size of the priority
// queue exceeds k
if (heap1.Count > K) {
heap1.Sort();
heap1.Reverse();
heap1.RemoveAt(0);
}
}
heap1.Sort();
heap1.Reverse();
// Print the k-th smallest element
Console.WriteLine(heap1[0]);
}
// Driver code
public static void Main(String[] args)
{
// Given array
int []vec = { 5, 20, 10, 7, 1 };
// Size of array
int N = vec.Length;
// Given K
int K = 2;
// Function Call
kthSmallest(vec, N, K % N);
}
}
// This code is contributed by gauravrajput1
java 描述语言
<script>
// Javascript program for the above approach
// Function to find kth smallest array element
function kthSmallest(v,N,K)
{
let heap1 = [];
for (let i = 0; i < N; ++i) {
// Insert elements into
// the priority queue
heap1.push(v[i]);
// If size of the priority
// queue exceeds k
if (heap1.length > K)
{
heap1.sort(function(a,b){
return a-b;
});
heap1.reverse();
heap1.shift();
}
}
heap1.sort(function(a,b){
return a-b;
});
heap1.reverse();
// Print the k-th smallest element
document.write(heap1[0] +"<br>");
}
// Driver code
// Given array
let vec=[5, 20, 10, 7, 1 ];
// Size of array
let N = vec.length;
// Given K
let K = 2;
// Function Call
kthSmallest(vec, N, K % N);
// This code is contributed by patel2127
</script>
Output:
5
时间复杂度: O(N LogK) 辅助空间: O(K)，因为优先级队列 ay 任何时候都保持在最大 K 个元素。
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