当元素在 1 到 n 范围内时，不同元素的总和

给定一个包含 n 个元素的数组，使得数组中的每个元素都是 1 到 n 范围内的整数，求该数组中所有不同元素的总和。

例子：

Input: arr[] = {5, 1, 2, 4, 6, 7, 3, 6, 7}
Output: 28
The distinct elements in the array are 1, 2, 
3, 4, 5, 6, 7

Input: arr[] = {1, 1, 1}
Output: 1

该问题已作为一般问题出现在这里，该解决方案也适用于上述情况。但是下面解释了一种更好的方法。
该方法是通过使那些索引处的元素为负来标记数组元素的出现。例如，a[0] = 1，a[1] = 1，a[2] = 1。
我们检查 a[abs(a[i])-1] 是否 >=0，如果是，则将 a[abs(a[i])-1] 标记为负数。即a[0] = 1 >=0，我们将a[1-1]标记为a[0] = -1。接下来，a[1]，检查(abs(a[1]-1)是否为+ve。如果是-ve，则表示a[1]之前已经出现过，否则为该元素的第一次出现。请参考以下代码。

C++

// C++ program to find sum of distinct elements
#include 
using namespace std;
 
// Returns sum of distinct elements in arr[] assuming
// that elements in a[] are in range from 1 to n.
int sumOfDistinct(int a[], int n)
{
    int sum = 0;
    for (int i = 0; i < n; i++) {
     
        // If element appears first time
        if (a[abs(a[i]) - 1] >= 0) {
            sum += abs(a[i]);
            a[abs(a[i]) - 1] *= -1;
        }
    }
    
    return sum;
}
 
// Driver code
int main()
{
    int a[] = { 5, 1, 2, 4, 6, 7, 3, 6, 7 };
    int n = sizeof(a)/sizeof(a[0]);
    cout << sumOfDistinct(a, n) << endl;
    return 0;
}

Java

// JAVA program to find sum of distinct
// elements in sorted order
import java.io.*;
import java.util.*;
import java.math.*;
 
class GFG{
     
    // Returns sum of distinct elements in arr[]
    // assuming that elements in a[] are in
    // range from 1 to n.
    static int sumOfDistinct(int a[], int n)
    {
        int sum = 0;
        for (int i = 0; i < n; i++) {
     
            // If element appears first time
            if (a[Math.abs(a[i]) - 1] >= 0) {
                sum += Math.abs(a[i]);
                a[Math.abs(a[i]) - 1] *= -1;
            }
        }
     
        return sum;
    }
 
 
    // Driver code
    public static void main(String args[])
    {
        int a[] = { 5, 1, 2, 4, 6, 7, 3, 6, 7 };
        int n = a.length;
        System.out.println(sumOfDistinct(a, n) );
    }
}
 
// This code is contributed by Nikita Tiwari.

Python3

# Python program to find sum of distinct elements
# in sorted order
import math
 
# Returns sum of distinct elements in arr[]
# assuming that elements in a[] are in
# range from 1 to n.
def sumOfDistinct(a , n) :
    sum = 0
    i = 0
    while i < n:
 
        # If element appears first time
        if (a[abs(a[i]) - 1] >= 0) :
            sum = sum + abs(a[i])
            a[abs(a[i]) - 1] = a[abs(a[i]) - 1] * (-1)
        i = i + 1
     
    return sum;
     
     
# Driver code
a = [ 5, 1, 2, 4, 6, 7, 3, 6, 7 ]
n = len(a)
print (sumOfDistinct(a, n))
     
# This code is contributed by Nikita Tiwari.

C#

// C# program to find sum of distinct
// elements in sorted order
using System;
 
class GFG{
     
    // Returns sum of distinct elements
    // in arr[] assuming that elements
    // in a[] are in range from 1 to n
    static int sumOfDistinct(int []a, int n)
    {
        int sum = 0;
        for (int i = 0; i < n; i++) {
     
            // If element appears first time
            if (a[Math.Abs(a[i]) - 1] >= 0) {
                sum += Math.Abs(a[i]);
                a[Math.Abs(a[i]) - 1] *= - 1;
            }
        }
     
        return sum;
    }
 
    // Driver code
    public static void Main()
    {
        int []a = {5, 1, 2, 4, 6, 7, 3, 6, 7};
        int n = a.Length;
        Console.Write(sumOfDistinct(a, n));
    }
}
 
// This code is contributed by Nitin Mittal

PHP

= 0)
        {
            $sum += abs($a[$i]);
            $a[abs($a[$i]) - 1] *= -1;
        }
    }
     
    return $sum;
}
 
    // Driver code
    $a = array(5, 1, 2, 4, 6, 7, 3, 6, 7);
    $n = sizeof($a);
    echo sumOfDistinct($a, $n) ;
 
// This code is contributed by nitin mittal
?>

Javascript

输出：