查找数组中非重复(不同)元素的总和

给定一个具有重复元素的整数数组，任务是查找数组中所有不同元素的总和。
例子：

Input  : arr[] = {12, 10, 9, 45, 2, 10, 10, 45,10};
Output : 78
Here we take 12, 10, 9, 45, 2 for sum
because it's distinct elements 

Input : arr[] = {1, 10, 9, 4, 2, 10, 10, 45 , 4};
Output : 71

一个简单的解决方案是使用两个嵌套循环。外循环从最左边的元素开始一个接一个地选择一个元素。内部循环检查元素的左侧是否存在。如果存在，则忽略该元素。
时间复杂度： O(n ² )
辅助空间： O(1)
这个问题的一个更好的解决方案是，使用排序技术，我们首先以升序对数组的所有元素进行排序，然后在数组中逐个查找不同的元素。

C++

// C++ Find the sum of all non-repeated
// elements in an array
#include
using namespace std;
 
// Find the sum of all non-repeated elements
// in an array
int findSum(int arr[], int n)
{
    // sort all elements of array
    sort(arr, arr + n);
 
    int sum = 0;
    for (int i=0; i

---

Java

import java.util.Arrays;
 
// Java Find the sum of all non-repeated
// elements in an array
public class GFG {
 
// Find the sum of all non-repeated elements
// in an array
    static int findSum(int arr[], int n) {
        // sort all elements of array
 
        Arrays.sort(arr);
        
        int sum = arr[0];
        for (int i = 0; i < n-1; i++) {
            if (arr[i] != arr[i + 1]) {
                sum = sum + arr[i+1];
            }
        }
 
        return sum;
    }
 
// Driver code
    public static void main(String[] args) {
        int arr[] = {1, 2, 3, 1, 1, 4, 5, 6};
        int n = arr.length;
        System.out.println(findSum(arr, n));
 
    }
}

Python3

# Python3 Find the sum of all non-repeated
# elements in an array
 
  
# Find the sum of all non-repeated elements
# in an array
def findSum(arr,  n):
    # sort all elements of array
    arr.sort()
  
    sum = arr[0]
    for i in range(0,n-1):
        if (arr[i] != arr[i+1]):
            sum = sum + arr[i+1]
     
    return sum
  
# Driver code
def main():
    arr= [1, 2, 3, 1, 1, 4, 5, 6]
    n = len(arr)
    print(findSum(arr, n))
 
if __name__ == '__main__':
    main()
# This code is contributed by 29AjayKumar

C#

// C# Find the sum of all non-repeated
// elements in an array
using System;
class GFG
{
 
    // Find the sum of all non-repeated elements
    // in an array
    static int findSum(int []arr, int n)
    {
        // sort all elements of array
        Array.Sort(arr);
         
        int sum = arr[0];
        for (int i = 0; i < n - 1; i++)
        {
            if (arr[i] != arr[i + 1])
            {
                sum = sum + arr[i + 1];
            }
        }
        return sum;
    }
 
    // Driver code
    public static void Main()
    {
        int []arr = {1, 2, 3, 1, 1, 4, 5, 6};
        int n = arr.Length;
        Console.WriteLine(findSum(arr, n));
    }
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// JavaScript Program to find the sum of all non-repeated
// elements in an array
 
// Find the sum of all non-repeated elements
// in an array
function findSum(arr, n)
{
    // sort all elements of array
    arr.sort();
 
    let sum = 0;
    for (let i=0; i<n; i++)
    {
        if (arr[i] != arr[i+1])
            sum = sum + arr[i];
    }
 
    return sum;
}
 
// Driver code
 
    let arr = [1, 2, 3, 1, 1, 4, 5, 6];
    let n = arr.length;
    document.write(findSum(arr, n));
 
// This code is contributed by Surbhi Tyagi
 

---

输出：

21

时间复杂度： O(n log n)
空间复杂度： O(1)
这个问题的有效解决方案是，使用unordered_set我们运行一个for循环，并且哪个值首次出现在其add sum变量中，并存储在哈希表中，下次我们不使用该值。

C++

// C++ Find the sum of all non- repeated
// elements in an array
#include
using namespace std;
 
// Find the sum of all non-repeated elements
// in an array
int findSum(int arr[],int n)
{
    int sum = 0;
 
    // Hash to store all element of array
    unordered_set< int > s;
    for (int i=0; i

Java

// Java Find the sum of all non- repeated
// elements in an array
import java.util.*;
 
class GFG
{
     
    // Find the sum of all non-repeated elements
    // in an array
    static int findSum(int arr[], int n)
    {
        int sum = 0;
 
        // Hash to store all element of array
        HashSet s = new HashSet();
        for (int i = 0; i < n; i++)
        {
            if (!s.contains(arr[i]))
            {
                sum += arr[i];
                s.add(arr[i]);
            }
        }
        return sum;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = {1, 2, 3, 1, 1, 4, 5, 6};
        int n = arr.length;
        System.out.println(findSum(arr, n));
    }
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 Find the sum of all
# non- repeated elements in an array
 
# Find the sum of all non-repeated
# elements in an array
def findSum(arr, n):
    s = set()
    sum = 0
 
    # Hash to store all element
    # of array
    for i in range(n):
        if arr[i] not in s:
            s.add(arr[i])
    for i in s:
        sum = sum + i
 
    return sum
 
# Driver code
arr = [1, 2, 3, 1, 1, 4, 5, 6]
n = len(arr)
print(findSum(arr, n))
 
# This code is contributed by Shrikant13

C＃

// C# Find the sum of all non- repeated
// elements in an array
using System;
using System.Collections.Generic;
 
class GFG
{
     
    // Find the sum of all non-repeated elements
    // in an array
    static int findSum(int []arr, int n)
    {
        int sum = 0;
 
        // Hash to store all element of array
        HashSet s = new HashSet();
        for (int i = 0; i < n; i++)
        {
            if (!s.Contains(arr[i]))
            {
                sum += arr[i];
                s.Add(arr[i]);
            }
        }
        return sum;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int []arr = {1, 2, 3, 1, 1, 4, 5, 6};
        int n = arr.Length;
        Console.WriteLine(findSum(arr, n));
    }
}
 
// This code is contributed by Rajput-Ji

输出：

21

时间复杂度： O(n)
辅助空间： O(n)

方法＃3：使用内置的Python函数：

方法：

• 使用Counter()函数计算频率
• 将频率键转换为列表。
• 计算列表的总和。

下面是上述方法的实现。

Python3

# Python program for the above approach
from collections import Counter
 
# Function to return the sum of distinct elements
def sumOfElements(arr, n):
 
    # Counter function is used to
    # calculate frequency of elements of array
    freq = Counter(arr)
     
    # Converting keys of freq dictionary to list
    lis = list(freq.keys())
     
    # Return sum of list
    return sum(lis)
 
 
# Driver code
if __name__ == "__main__":
 
    arr = [1, 2, 3, 1, 1, 4, 5, 6]
    n = len(arr)
 
    print(sumOfElements(arr, n))
 
# This code is contributed by vikkycirus

输出：

21