数组中所有偶数出现的元素的总和

给定一个包含重复元素的整数数组。任务是找到给定数组中所有偶数出现的元素的总和。那是数组中频率为偶数的所有此类元素的总和。

例子：

Input : arr[] = {1, 1, 2, 2, 3, 3, 3}
Output : 6
The even occurring element are 1 and 2 (both occur two times). 
Therefore sum of all 1's in the array = 1+1+2+2 = 6.

Input : arr[] = {10, 20, 30, 40, 40}
Output : 80
Element with even frequency are 40.
Sum = 40.

方法：

• 遍历数组并使用C++中的unordered_map来存储数组元素的频率，这样map的key就是数组元素，value就是它在数组中的频率。
• 然后，遍历map，查找元素出现的频率并检查是否为偶数，如果为偶数，则将此元素添加到sum中。

下面是上述方法的实现：

C++

// C++ program to find the sum of all even
// occurring elements in an array
 
#include 
using namespace std;
 
// Function to find the sum of all even
// occurring elements in an array
int findSum(int arr[], int N)
{
    // Map to store frequency of elements
    // of the array
    unordered_map mp;
 
    for (int i = 0; i < N; i++) {
        mp[arr[i]]++;
    }
 
    // variable to store sum of all
    // even occurring elements
    int sum = 0;
 
    // loop to iterate through map
    for (auto itr = mp.begin(); itr != mp.end(); itr++) {
 
        // check if frequency is even
        if (itr->second % 2 == 0)
            sum += (itr->first) * (itr->second);
    }
 
    return sum;
}
 
// Driver Code
int main()
{
    int arr[] = { 10, 20, 20, 40, 40 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << findSum(arr, N);
 
    return 0;
}

Java

// Java program to find the sum of all even
// occurring elements in an array
 
import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;
import java.util.Map.Entry;
 
public class GFG {
    public static int element(int[] arr, int n)
    {
 
        Map map = new HashMap();
        for (int i = 0; i < n; i++) {
            int count = 0;
            if (map.get(arr[i]) != null) {
                count = map.get(arr[i]);
            }
            map.put(arr[i], count + 1);
        }
 
        int sum = 0;
        for (Entry entry : map.entrySet()) {
            if (entry.getValue() % 2 == 0) {
                sum += entry.getKey() * entry.getValue();
            }
        }
 
        return sum;
    }
 
    public static void main(String[] args)
    {
        int arr[] = { 1, 1, 2, 2, 3, 3, 3 };
 
        // sum should be = 1+1+2+2=6;
        int n = arr.length;
        System.out.println(element(arr, n));
    }
}

Python3

# Python3 program to find the sum
# of all even occurring elements
# in an array
 
# Function to find the sum of all even
# occurring elements in an array
def findSum(arr, N):
 
    # Map to store frequency of
    # elements of the array
    mp = {}
 
    for i in range(N):
        if arr[i] in mp:
            mp[arr[i]] += 1
        else:
            mp[arr[i]] = 1
 
    # Variable to store sum of all
    # even occurring elements
    Sum = 0
 
    # Loop to iterate through map
    for first, second in mp.items():
 
        # Check if frequency is even
        if (second % 2 == 0):
            Sum += (first) * (second)
 
    return Sum
     
# Driver code   
arr = [ 10, 20, 20, 40, 40 ]
 
N = len(arr)
 
print(findSum(arr, N))
 
# This code is contributed by divyeshrabadiya07

C#

// C# program to find the sum of all even
// occurring elements in an array
using System;
using System.Collections.Generic;
 
class GFG {
     
    static int element(int[] arr, int n)
    {
        Dictionary map = new Dictionary(); 
        for (int i = 0; i < n; i++)
        {
             
            if(!map.ContainsKey(arr[i]))
            {
                map.Add(arr[i], 1);
            }
            else
            {
                map[arr[i]] += 1;
            }
             
        }
  
        int sum = 0;
         
        foreach(KeyValuePair entry in map)
        {
            if (entry.Value % 2 == 0)
            {
                sum += entry.Key * entry.Value;
            }
        }
  
        return sum;
    }
   
  // Driver code
  static void Main() {
        int[] arr = { 10, 20, 20, 40, 40};
  
        int n = arr.Length;
        Console.WriteLine(element(arr, n));
  }
}
 
// This code is contributed by divyesh072019

Javascript

// Javascript program to find the sum of all odd
// occurring elements in an array
 
// Function to find the sum of all odd
// occurring elements in an array
function findSum(arr, N)
{
 
    // Store frequencies of elements
    // of the array
    let mp = new Map();
    for (let i = 0; i < N; i++) {
        if (mp.has(arr[i])) {
            mp.set(arr[i], mp.get(arr[i]) + 1)
        } else {
            mp.set(arr[i], 1)
        }
    }
 
    // variable to store sum of all
    // odd occurring elements
    let sum = 0;
 
    // loop to iterate through map
    for (let itr of mp) {
 
        // check if frequency is odd
        if (itr[1] % 2 == 0)
            sum += (itr[0]) * (itr[1]);
    }
 
    return sum;
}
 
// Driver Code
 
let arr = [10, 20, 20, 40, 40];
 
let N = arr.length
 
document.write(findSum(arr, N));
 
// This code is contributed by _saurabh_jaiswal.

输出：

120

时间复杂度：O(N)，其中 N 是数组中元素的数量。

方法2：使用内置Python函数：

• 使用 Counter函数计算每个元素的频率
• 遍历频率字典并求和所有出现偶数与其频率相乘的元素。

下面是实现：

蟒蛇3

# Python3 implementation
from collections import Counter
 
 
def sumEven(arr, n):
 
    # Counting frequency of every
    # element using Counter
    freq = Counter(arr)
     
    # initializing sum 0
    sum = 0
     
    # Traverse the freq and print all
    # sum all elements with even frequency
    # multiplied by its frequency
    for it in freq:
        if freq[it] % 2 == 0:
            sum = sum + it*freq[it]
    print(sum)
 
 
# Driver code
arr = [10, 20, 20, 40, 40]
n = len(arr)
 
sumEven(arr, n)
 
# This code is contributed by vikkycirus

输出：

120

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