查找数组中所有重复元素的频率

给定一个包含重复元素的整数数组，任务是找到数组中的重复元素及其频率。

例子：

Input: arr[] = {2, 3, 4, 5, 4, 6, 4, 7, 4, 5, 6, 6}
Output: Below is the frequency of repeated elements –
4 –> 4
5 –> 2
6 –> 3

Input: arr[] = {4, 4, 5, 5, 6}
Output: Below is the frequency of repeated elements –
4 –> 2
5 –> 2

方法：

• 创建一个哈希映射来存储元素的频率。
• 频率大于1的元素为重复元素。

下面是上述方法的实现：

CPP

// CPP Implementation to find the
// repeating elements with there count
#include
using namespace std;
  
// Function to find the repeating
// elements with there count
map findRepeating(int arr[], int size){
  
    // Hash map to store the 
    // frequency of elements
    map frequency;
      
    // Loop to store the frequency of 
    // elements of array
    for (int i = 0; i < size; i++)
    frequency[arr[i]]++;
    return frequency;
}
          
// Driver Code 
int main(){
    int arr[] = {4, 4, 5, 5, 6};
    int arr_size = sizeof(arr)/sizeof(arr[0]); 
    map frequency = findRepeating(arr, arr_size);
    cout<<"Below is the frequency of repeated elements -"< 1)
            cout< "<

---

Java

// Java Implementation to find the
// repeating elements with there count
import java.util.*;
  
class GFG
{
      
// Function to find the repeating
// elements with there count
static HashMap findRepeating(int []arr, int size){
      
    // Hash map to store the 
    // frequency of elements
    HashMap frequency = new HashMap();
      
    // Loop to store the frequency of 
    // elements of array
    for(int i = 0; i < size; i++) 
    {
        if(frequency.containsKey(arr[i]))
        {
            frequency.put(arr[i], frequency.get(arr[i]) + 1);
        }
        else
        {
            frequency.put(arr[i], 1);
        }
    }
    return frequency;
} 
  
// Driver Code 
public static void main(String []args)
{
    int []arr = {4, 4, 5, 5, 6};
    int arr_size = arr.length; 
    HashMap frequency = findRepeating(arr, arr_size);
    System.out.println("Below is the frequency"
    +"of repeated elements -");
    for (Map.Entry entry : frequency.entrySet())
        if (entry.getValue() > 1)
            System.out.println(entry.getKey()+ " --> "+entry.getValue());
}
}
  
// This code is contributed by PrinciRaj1992

Python

# Python Implementation to find the
# repeating elements with there count
  
# Function to find the repeating
# elements with there count
def findRepeating(arr, size):
      
    # Hash map to store the 
    # frequency of elements
    frequency = {}
      
    # Loop to store the frequency of 
    # elements of array
    for i in range (0, size):
        frequency[arr[i]] = \
        frequency.get(arr[i], 0) + 1
    return frequency
      
# Driver Code 
if __name__ == "__main__":
    arr = [4, 4, 5, 5, 6]
    arr_size = len(arr) 
    frequency = findRepeating(arr, arr_size)
    print("Below is the frequency\
    of repeated elements -")
    for i in frequency:
        if frequency[i] > 1:
            print(i, " --> ", frequency[i])

C#

// C# Implementation to find the
// repeating elements with there count
using System;
using System.Collections.Generic;
  
class GFG
{
      
// Function to find the repeating
// elements with there count
static Dictionary findRepeating(int []arr, int size){
      
    // Hash map to store the 
    // frequency of elements
    Dictionary frequency = new Dictionary();
      
    // Loop to store the frequency of 
    // elements of array
    for(int i = 0; i < size; i++) 
    {
        if(frequency.ContainsKey(arr[i]))
        {
            frequency[arr[i]] = frequency[arr[i]] + 1;
        }
        else
        {
            frequency.Add(arr[i], 1);
        }
    }
    return frequency;
} 
  
// Driver Code 
public static void Main(String []args)
{
    int []arr = {4, 4, 5, 5, 6};
    int arr_size = arr.Length; 
    Dictionary frequency = findRepeating(arr, arr_size);
    Console.WriteLine("Below is the frequency"
                        +"of repeated elements -");
    foreach (KeyValuePair entry in frequency)
        if (entry.Value > 1)
            Console.WriteLine(entry.Key+ " --> "+entry.Value);
}
}
  
// This code is contributed by 29AjayKumar

输出：

Below is the frequency    of repeated elements -
4  -->  2
5  -->  2

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