每次插入后第 K 个最大元素

给定一个无限的整数流，找出任意时间点的第 k 个最大元素。可以假设 1 <= k <= n。

Input:
stream[] = {10, 20, 11, 70, 50, 40, 100, 5, ...}
k = 3
Output:    {_,   _, 10, 11, 20, 40, 50,  50, ...}

允许的额外空间为 O(k)。

这个想法是使用最小堆。
1) 在最小堆中存储前 k 个元素。
2) 对于从 (k+1)-th 到第 n-th 的每个元素，请执行以下操作。
……a) 打印堆的根。
……b) 如果当前元素大于堆的根，则弹出根并插入

CPP

// CPP program to find k-th largest element in a
// stream after every insertion.
#include 
using namespace std;
 
int kthLargest(int stream[], int n, int k)
{
   // Create a min heap and store first k-1 elements
   // of stream into
   priority_queue, greater > pq;
 
   // Push first k elements and print "_" (k-1) times
   for (int i=0; i pq.top())
       {
           pq.pop();
           pq.push(stream[i]);
       } 
   } 
 
   // Print last k-th largest element (after
   // (inserting last element)
   cout << pq.top();
}
 
// Driver code
int main()
{
   int arr[] = {10, 20, 11, 70, 50, 40, 100, 55};
   int k = 3;
   int n = sizeof(arr)/sizeof(arr[0]);
   kthLargest(arr, n, k);
   return 0;
}

Java

// Java Program for the above approach
import java.util.*;
class GFG
{
  static void kthLargest(int stream[], int n, int k)
  {
 
    // Create a min heap and store first k-1 elements
    // of stream into
    Vector pq = new Vector(n);
 
    // Push first k elements and print "_" (k-1) times
    for (int i = 0; i < k - 1; i++)
    {
      pq.add(stream[i]);
      System.out.print("_ ");
    }
    pq.add(stream[k - 1]);
 
    for (int i = k; i < n; i++)
    {
 
      // We must insert last element before we
      // decide last k-th largest output.
      Collections.sort(pq);
      System.out.print(pq.get(0) + " ");     
      if (stream[i] > pq.get(0))
      {
        pq.remove(0);
        pq.add(stream[i]);
      }  
    }  
 
    // Print last k-th largest element (after
    // (inserting last element)
    Collections.sort(pq);
    System.out.print(pq.get(0));
  }
 
  // Driver code
  public static void main(String[] args) {
    int arr[] = {10, 20, 11, 70, 50, 40, 100, 55};
    int k = 3;
    int n = arr.length;
    kthLargest(arr, n, k);
  }
}
 
// This code is contributed by divyeshrabadiya07.

Python3

# Python Program for the above approach
def kthLargest(stream, n, k):
     
    # Create a min heap and store first k-1 elements
    # of stream into
    pq = []
     
    # Push first k elements and print "_" (k-1) times
    for i in range(k - 1):
        pq.append(stream[i])
        print("_ ", end = "")
    pq.append(stream[k - 1])   
    for i in range(k, n):
       
        # We must insert last element before we
        # decide last k-th largest output.
        pq.sort()
        print(pq[0], end = " ")
        if(stream[i] > pq[0]):
            del pq[0]
            pq.append(stream[i])
     
    # Print last k-th largest element (after
    # (inserting last element)
    pq.sort()
    print(pq[0], end = "")
 
# Driver code
arr = [10, 20, 11, 70, 50, 40, 100, 55]
k = 3
n = len(arr)
kthLargest(arr, n, k)
 
# This code is contributed by avanitrachhadiya2155

C#

// C# program to find k-th largest element in a 
// stream after every insertion.
using System;
using System.Collections.Generic;
class GFG
{
  static void kthLargest(int[] stream, int n, int k)
  {
 
    // Create a min heap and store first k-1 elements
    // of stream into
    List pq = new List();
 
    // Push first k elements and print "_" (k-1) times
    for (int i = 0; i < k - 1; i++)
    {
      pq.Add(stream[i]);
      Console.Write("_ ");
    }
    pq.Add(stream[k - 1]);
 
    for (int i = k; i < n; i++)
    {
 
      // We must insert last element before we
      // decide last k-th largest output.
      pq.Sort();
      Console.Write(pq[0] + " ");     
      if (stream[i] > pq[0])
      {
        pq.RemoveAt(0);
        pq.Add(stream[i]);
      }  
    }  
 
    // Print last k-th largest element (after
    // (inserting last element)
    pq.Sort();
    Console.Write(pq[0]);
  }
 
  // Driver code
  static void Main()
  {
    int[] arr = {10, 20, 11, 70, 50, 40, 100, 55};
    int k = 3;
    int n = arr.Length;
    kthLargest(arr, n, k);
  }
}
 
// This code is contributed by divyesh072019.

Javascript

 
// JavaScript program to find k-th largest element in a
// stream after every insertion.
 
function kthLargest(stream, n, k) {
 
    // Create a min heap and store first k-1 elements
    // of stream into
    let pq = new Array();
 
    // Push first k elements and print "_" (k-1) times
    for (let i = 0; i < k - 1; i++) {
        pq.push(stream[i]);
        document.write("_ ");
    }
    pq.push(stream[k - 1]);
 
    for (let i = k; i < n; i++) {
 
        // We must insert last element before we
        // decide last k-th largest output.
        pq.sort((a, b) => a - b);
        document.write(pq[0] + " ");
        if (stream[i] > pq[0]) {
            pq.shift(0);
            pq.push(stream[i]);
        }
    }
 
    // Print last k-th largest element (after
    // (inserting last element)
    pq.sort((a, b) => a - b);
    document.write(pq[0]);
}
 
// Driver code
 
let arr = [10, 20, 11, 70, 50, 40, 100, 55];
let k = 3;
let n = arr.length;
kthLargest(arr, n, k);
 
// This code is contributed by _saurabh_jaiswal
 

输出：

_ _ 10 11 20 40 50 55

如果流包含非原始类型的元素，我们可以定义我们自己的压缩器函数并相应地创建一个 priority_queue。

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