查找出现b次的唯一元素
原文：https://www.geeksforgeeks.org/find-element-appears-b-times/
给定一个数组，其中每个元素出现一次，除了一个元素出现b（a > b）次。 找到发生b次的元素。
示例：
Input : arr[] = [1, 1, 2, 2, 2, 3, 3, 3]
a = 3, b = 2
Output : 1
将每个数字相加一次，然后将总和乘以a，我们将得到数组每个元素之和的乘积。 将其存储为a_sum。 从a_sum中减去整个数组的总和，然后将结果除以a - b。 我们得到的数字是所需的数字（在数组中出现b次）。
C++
// CPP program to find the only element that
// appears b times
#include <bits/stdc++.h>
using namespace std;
int appearsbTimes(int arr[], int n, int a, int b)
{
unordered_set<int> s;
int a_sum = 0, sum = 0;
for (int i = 0; i < n; i++) {
if (s.find(arr[i]) == s.end()) {
s.insert(arr[i]);
a_sum += arr[i];
}
sum += arr[i];
}
a_sum = a * a_sum;
return ((a_sum - sum) / (a - b));
}
int main()
{
int arr[] = { 1, 1, 2, 2, 2, 3, 3, 3 };
int a = 3, b = 2;
int n = sizeof(arr) / sizeof(arr[0]);
cout << appearsbTimes(arr, n, a, b);
return 0;
}
Java
// Java program to find the only element that
// appears b times
import java.util.*;
class GFG
{
static int appearsbTimes(int arr[], int n,
int a, int b)
{
HashSet<Integer> s = new HashSet<Integer>();
int a_sum = 0, sum = 0;
for (int i = 0; i < n; i++)
{
if (!s.contains(arr[i]))
{
s.add(arr[i]);
a_sum += arr[i];
}
sum += arr[i];
}
a_sum = a * a_sum;
return ((a_sum - sum) / (a - b));
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 1, 1, 2, 2, 2, 3, 3, 3 };
int a = 3, b = 2;
int n = arr.length;
System.out.println(appearsbTimes(arr, n, a, b));
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 program to find the only
# element that appears b times
def appearsbTimes(arr, n, a, b):
s = dict()
a_Sum = 0
Sum = 0
for i in range(n):
if (arr[i] not in s.keys()):
s[arr[i]] = 1
a_Sum += arr[i]
Sum += arr[i]
a_Sum = a * a_Sum
return ((a_Sum - Sum) // (a - b))
# Driver code
arr = [1, 1, 2, 2, 2, 3, 3, 3]
a, b = 3, 2
n = len(arr)
print(appearsbTimes(arr, n, a, b))
# This code is contributed by mohit kumar
C
// C# program to find the only element that
// appears b times
using System;
using System.Collections.Generic;
class GFG
{
static int appearsbTimes(int []arr, int n,
int a, int b)
{
HashSet<int> s = new HashSet<int>();
int a_sum = 0, sum = 0;
for (int i = 0; i < n; i++)
{
if (!s.Contains(arr[i]))
{
s.Add(arr[i]);
a_sum += arr[i];
}
sum += arr[i];
}
a_sum = a * a_sum;
return ((a_sum - sum) / (a - b));
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 1, 1, 2, 2, 2, 3, 3, 3 };
int a = 3, b = 2;
int n = arr.Length;
Console.WriteLine(appearsbTimes(arr, n, a, b));
}
}
// This code is contributed by Princi Singh
输出：
1
请参考下面的文章以了解更多方法。
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