查找数组中非重复（不同）元素的总和
原文： https://www.geeksforgeeks.org/find-sum-non-repeating-distinct-elements-array/
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。
示例：
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
简单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)。
这个问题的更好的解决方案是，使用排序技术，我们首先按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<bits/stdc++.h>
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<n; i++)
{
if (arr[i] != arr[i+1])
sum = sum + arr[i];
}
return sum;
}
// Driver code
int main()
{
int arr[] = {1, 2, 3, 1, 1, 4, 5, 6};
int n = sizeof(arr)/sizeof(int);
cout << findSum(arr, n);
return 0;
}
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
输出：
21
时间复杂度：O(n log n)。
空间复杂度：O(1)。
此问题的有效解决方案是，使用unordered_set我们运行一个for循环，并且该值首次出现在其add sum变量中，并存储在哈希表中，下一次我们不使用该值。
C++
// C++ Find the sum of all non- repeated
// elements in an array
#include<bits/stdc++.h>
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<n; i++)
{
if (s.find(arr[i]) == s.end())
{
sum += arr[i];
s.insert(arr[i]);
}
}
return sum;
}
// Driver code
int main()
{
int arr[] = {1, 2, 3, 1, 1, 4, 5, 6};
int n = sizeof(arr)/sizeof(int);
cout << findSum(arr, n);
return 0;
}
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<Integer> s = new HashSet<Integer>();
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<int> s = new HashSet<int>();
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)。
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