计算数组中的不同元素

给定一个未排序的数组，请计算其中所有不同的元素。
例子：

Input :   arr[] = {10, 20, 20, 10, 30, 10}
Output : 3
There are three distinct elements 10, 20 and 30.

Input :   arr[] = {10, 20, 20, 10, 20}
Output : 2

一个简单的解决方案是运行两个循环。对于每个元素，检查它是否已经出现过。如果是，则增加不同元素的计数。

C++

// C++ program to count distinct elements
// in a given array
#include 
using namespace std;
 
int countDistinct(int arr[], int n)
{
    int res = 1;
 
    // Pick all elements one by one
    for (int i = 1; i < n; i++) {
        int j = 0;
        for (j = 0; j < i; j++)
            if (arr[i] == arr[j])
                break;
 
        // If not printed earlier, then print it
        if (i == j)
            res++;
    }
    return res;
}
 
// Driver program to test above function
int main()
{
    int arr[] = { 12, 10, 9, 45, 2, 10, 10, 45 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << countDistinct(arr, n);
    return 0;
}

Java

// Java program to count distinct
// elements in a given array
class GFG
{
     
static int countDistinct(int arr[], int n)
{
    int res = 1;
 
    // Pick all elements one by one
    for (int i = 1; i < n; i++)
    {
        int j = 0;
        for (j = 0; j < i; j++)
            if (arr[i] == arr[j])
                break;
 
        // If not printed earlier,
        // then print it
        if (i == j)
            res++;
    }
    return res;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 12, 10, 9, 45, 2, 10, 10, 45 };
    int n = arr.length;
    System.out.println(countDistinct(arr, n));
}
}
 
// This code is contributed by Code_Mech.

Python3

# Python3 program to count distinct
# elements in a given array
def countDistinct(arr, n):
 
    res = 1
 
    # Pick all elements one by one
    for i in range(1, n):
        j = 0
        for j in range(i):
            if (arr[i] == arr[j]):
                break
 
        # If not printed earlier, then print it
        if (i == j + 1):
            res += 1
     
    return res
 
# Driver Code
arr = [12, 10, 9, 45, 2, 10, 10, 45]
n = len(arr)
print(countDistinct(arr, n))
 
# This code is contributed by Mohit Kumar

C#

// C# program to count distinct
// elements in a given array
using System;
 
class GFG
{
     
static int countDistinct(int []arr, int n)
{
    int res = 1;
 
    // Pick all elements one by one
    for (int i = 1; i < n; i++)
    {
        int j = 0;
        for (j = 0; j < i; j++)
            if (arr[i] == arr[j])
                break;
 
        // If not printed earlier,
        // then print it
        if (i == j)
            res++;
    }
    return res;
}
 
// Driver code
public static void Main()
{
    int []arr = { 12, 10, 9, 45,
                  2, 10, 10, 45 };
    int n = arr.Length;
    Console.WriteLine(countDistinct(arr, n));
}
}
 
// This code is contributed
// by Akanksha Rai

Javascript

 
// JavaScript program to count distinct elements
// in a given array
 
function countDistinct(arr, n)
{
    let res = 1;
 
    // Pick all elements one by one
    for (let i = 1; i < n; i++) {
        let j = 0;
        for (j = 0; j < i; j++)
            if (arr[i] === arr[j])
                break;
 
        // If not printed earlier, then print it
        if (i === j)
            res++;
    }
    return res;
}
 
// Driver program to test above function
    let arr = [ 12, 10, 9, 45, 2, 10, 10, 45 ];
    let n = arr.length;
    document.write(countDistinct(arr, n));
 
 
// This code is contributed by Surbhi Tyagi
 

输出：

5

上述解决方案的时间复杂度为O(n ² )。我们可以使用排序来解决O(nLogn)时间中的问题。这个想法很简单，首先对数组进行排序，以便每个元素的所有出现变为连续。一旦出现连续，我们就可以遍历排序后的数组，并在O(n)时间内对不同元素进行计数。以下是该想法的实施。

C++

// C++ program to count all distinct elements
// in a given array
#include 
#include 
using namespace std;
 
int countDistinct(int arr[], int n)
{
    // First sort the array so that all
    // occurrences become consecutive
    sort(arr, arr + n);
 
    // Traverse the sorted array
    int res = 0;
    for (int i = 0; i < n; i++) {
 
        // Move the index ahead while
        // there are duplicates
        while (i < n - 1 && arr[i] == arr[i + 1])
            i++;
 
        res++;
    }
 
    return res;
}
 
// Driver program to test above function
int main()
{
    int arr[] = { 6, 10, 5, 4, 9, 120, 4, 6, 10 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << countDistinct(arr, n);
    return 0;
}

Java

// Java program to count all distinct elements
// in a given array
import java.util.Arrays;
 
class GFG
{
 
    static int countDistinct(int arr[], int n)
    {
        // First sort the array so that all
        // occurrences become consecutive
        Arrays.sort(arr);
 
        // Traverse the sorted array
        int res = 0;
        for (int i = 0; i < n; i++)
        {
 
            // Move the index ahead while
            // there are duplicates
            while (i < n - 1 &&
                    arr[i] == arr[i + 1])
            {
                i++;
            }
            res++;
        }
        return res;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = {6, 10, 5, 4, 9, 120, 4, 6, 10};
        int n = arr.length;
        System.out.println(countDistinct(arr, n));
    }
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to count all distinct
# elements in a given array
 
def countDistinct(arr, n):
 
    # First sort the array so that all
    # occurrences become consecutive
    arr.sort();
 
    # Traverse the sorted array
    res = 0;
    i = 0;
    while(i < n):
 
        # Move the index ahead while
        # there are duplicates
        while (i < n - 1 and
               arr[i] == arr[i + 1]):
            i += 1;
 
        res += 1;
        i += 1;
 
    return res;
 
# Driver Code
arr = [ 6, 10, 5, 4, 9, 120, 4, 6, 10 ];
n = len(arr);
print(countDistinct(arr, n));
 
# This code is contributed by mits

C＃

// C# program to count all distinct elements
// in a given array
using System;
 
class GFG
{
 
    static int countDistinct(int[] arr, int n)
    {
        // First sort the array so that all
        // occurrences become consecutive
        Array.Sort(arr);
 
        // Traverse the sorted array
        int res = 0;
        for (int i = 0; i < n; i++)
        {
 
            // Move the index ahead while
            // there are duplicates
            while (i < n - 1 &&
                    arr[i] == arr[i + 1])
            {
                i++;
            }
            res++;
        }
        return res;
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = {6, 10, 5, 4, 9, 120, 4, 6, 10};
        int n = arr.Length;
        Console.WriteLine(countDistinct(arr, n));
    }
}
 
// This code is contributed by Code_Mech.

输出：

6

我们可以使用散列来平均解决O(n)时间。这个想法是从左到右遍历给定的数组，并跟踪哈希集中的访问元素，因为一个哈希集仅包含唯一元素。
以下是该想法的实施。

C++

/* CPP program to print all distinct elements
   of a given array */
#include 
using namespace std;
 
// This function prints all distinct elements
int countDistinct(int arr[], int n)
{
    // Creates an empty hashset
    unordered_set s;
 
    // Traverse the input array
    int res = 0;
    for (int i = 0; i < n; i++) {
 
        // If not present, then put it in
        // hashtable and increment result
        if (s.find(arr[i]) == s.end()) {
            s.insert(arr[i]);
            res++;
        }
    }
 
    return res;
}
 
// Driver program to test above function
int main()
{
    int arr[] = { 6, 10, 5, 4, 9, 120, 4, 6, 10 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << countDistinct(arr, n);
    return 0;
}

Java

// Java Program to count
// Unique elements in Array
import java.util.*;
 
class GFG
{
     
    // This method returns count
    // of Unique elements
    public static int countDistinct(int arr[],int n)
    {
 
        HashSet hs = new HashSet();
 
        for(int i = 0; i < n; i++)
        {
            // add all the elements to the HashSet
            hs.add(arr[i]);
        }
         
        // return the size of hashset as
        // it consists of all Unique elements
        return hs.size();    
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = new int[]{6, 10, 5, 4, 9,
                                120, 4, 6, 10};
        System.out.println(countDistinct(arr,
                                arr.length));
    }
}
 
// This code is contributed by Adarsh_Verma

Python3

''' Python3 program to prall distinct elements
of a given array '''
 
# This function prints all distinct elements
def countDistinct(arr, n):
 
    # Creates an empty hashset
    s = set()
     
    # Traverse the input array
    res = 0
    for i in range(n):
     
        # If not present, then put it in
        # hashtable and increment result
        if (arr[i] not in s):
            s.add(arr[i])
            res += 1
     
    return res
 
# Driver code
arr = [6, 10, 5, 4, 9, 120, 4, 6, 10]
n = len(arr)
print(countDistinct(arr, n))
 
# This code is contributed by SHUBHAMSINGH10

C＃

// C# Program to count
// Unique elements in Array
using System;
using System.Collections.Generic;
 
class GFG
{
     
    // This method returns count
    // of Unique elements
    public static int countDistinct(int []arr,int n)
    {
 
        HashSet hs = new HashSet();
 
        for(int i = 0; i < n; i++)
        {
            // add all the elements to the HashSet
            hs.Add(arr[i]);
        }
         
        // return the size of hashset as
        // it consists of all Unique elements
        return hs.Count;    
    }
 
    // Driver code
    public static void Main()
    {
        int []arr = new int[]{6, 10, 5, 4, 9,
                                120, 4, 6, 10};
        Console.WriteLine(countDistinct(arr,
                                arr.Length));
    }
}
 
/* This code contributed by PrinciRaj1992 */

输出：

6