数组中最常出现的元素

给定一个数组，在其中找到最频繁的元素。如果有多个元素出现的次数最多，请打印其中的任何一个。
例子：

Input : arr[] = {1, 3, 2, 1, 4, 1}
Output : 1
1 appears three times in array which
is maximum frequency.

Input : arr[] = {10, 20, 10, 20, 30, 20, 20}
Output : 20

一个简单的解决方案是运行两个循环。外循环一一挑选所有元素。内循环找到所拾取元素的频率，并与迄今为止的最大值进行比较。该解决方案的时间复杂度为O(n ² )
更好的解决方案是进行排序。我们首先对数组进行排序，然后线性遍历数组。

C++

// CPP program to find the most frequent element
// in an array.
#include 
using namespace std;
 
int mostFrequent(int arr[], int n)
{
    // Sort the array
    sort(arr, arr + n);
 
    // find the max frequency using linear traversal
    int max_count = 1, res = arr[0], curr_count = 1;
    for (int i = 1; i < n; i++) {
        if (arr[i] == arr[i - 1])
            curr_count++;
        else {
            if (curr_count > max_count) {
                max_count = curr_count;
                res = arr[i - 1];
            }
            curr_count = 1;
        }
    }
 
    // If last element is most frequent
    if (curr_count > max_count)
    {
        max_count = curr_count;
        res = arr[n - 1];
    }
 
    return res;
}
 
// driver program
int main()
{
    int arr[] = { 1, 5, 2, 1, 3, 2, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << mostFrequent(arr, n);
    return 0;
}

Java

//Java program to find the most frequent element
//in an array
import java.util.*;
 
class GFG {
     
    static int mostFrequent(int arr[], int n)
    {
         
        // Sort the array
        Arrays.sort(arr);
         
        // find the max frequency using linear
        // traversal
        int max_count = 1, res = arr[0];
        int curr_count = 1;
         
        for (int i = 1; i < n; i++)
        {
            if (arr[i] == arr[i - 1])
                curr_count++;
            else
            {
                if (curr_count > max_count)
                {
                    max_count = curr_count;
                    res = arr[i - 1];
                }
                curr_count = 1;
            }
        }
     
        // If last element is most frequent
        if (curr_count > max_count)
        {
            max_count = curr_count;
            res = arr[n - 1];
        }
     
        return res;
    }
     
    // Driver program
    public static void main (String[] args) {
         
        int arr[] = {1, 5, 2, 1, 3, 2, 1};
        int n = arr.length;
         
        System.out.println(mostFrequent(arr,n));
         
    }
}
 
// This code is contributed by Akash Singh.

Python3

# Python3 program to find the most
# frequent element in an array.
 
def mostFrequent(arr, n):
 
    # Sort the array
    arr.sort()
 
    # find the max frequency using
    # linear traversal
    max_count = 1; res = arr[0]; curr_count = 1
     
    for i in range(1, n):
        if (arr[i] == arr[i - 1]):
            curr_count += 1
             
        else :
            if (curr_count > max_count):
                max_count = curr_count
                res = arr[i - 1]
             
            curr_count = 1
     
    # If last element is most frequent
    if (curr_count > max_count):
     
        max_count = curr_count
        res = arr[n - 1]
     
    return res
 
# Driver Code
arr = [1, 5, 2, 1, 3, 2, 1]
n = len(arr)
print(mostFrequent(arr, n))
 
# This code is contributed by Smitha Dinesh Semwal.

C#

// C# program to find the most
// frequent element in an array
using System;
 
class GFG {
     
    static int mostFrequent(int []arr, int n)
    {
         
        // Sort the array
        Array.Sort(arr);
         
        // find the max frequency using
        // linear traversal
        int max_count = 1, res = arr[0];
        int curr_count = 1;
         
        for (int i = 1; i < n; i++)
        {
            if (arr[i] == arr[i - 1])
                curr_count++;
            else
            {
                if (curr_count > max_count)
                {
                    max_count = curr_count;
                    res = arr[i - 1];
                }
                curr_count = 1;
            }
        }
     
        // If last element is most frequent
        if (curr_count > max_count)
        {
            max_count = curr_count;
            res = arr[n - 1];
        }
     
        return res;
    }
     
    // Driver code
    public static void Main ()
    {
         
        int []arr = {1, 5, 2, 1, 3, 2, 1};
        int n = arr.Length;
         
        Console.WriteLine(mostFrequent(arr,n));
         
    }
}
 
// This code is contributed by vt_m.

PHP

 $max_count)
            {
                $max_count = $curr_count;
                $res = $arr[$i - 1];
            }
            $curr_count = 1;
        }
    }
 
    // If last element
    // is most frequent
    if ($curr_count > $max_count)
    {
        $max_count = $curr_count;
        $res = $arr[$n - 1];
    }
 
    return $res;
}
 
// Driver Code
{
    $arr = array(1, 5, 2, 1, 3, 2, 1);
    $n = sizeof($arr) / sizeof($arr[0]);
    echo mostFrequent($arr, $n);
    return 0;
}
 
// This code is contributed by nitin mittal
?>

Javascript

// JavaScript program to find the most frequent element
//in an array
 
    function mostFrequent(arr, n)
    {
           
        // Sort the array
        arr.sort();
           
        // find the max frequency using linear
        // traversal
        let max_count = 1, res = arr[0];
        let curr_count = 1;
           
        for (let i = 1; i < n; i++)
        {
            if (arr[i] == arr[i - 1])
                curr_count++;
            else
            {
                if (curr_count > max_count)
                {
                    max_count = curr_count;
                    res = arr[i - 1];
                }
                curr_count = 1;
            }
        }
       
        // If last element is most frequent
        if (curr_count > max_count)
        {
            max_count = curr_count;
            res = arr[n - 1];
        }
        return res;
    }
       
// Driver Code
 
        let arr = [1, 5, 2, 1, 3, 2, 1];
        let n = arr.length;
        document.write(mostFrequent(arr,n));
 
// This code is contributed by code_hunt.

输出 ：

1

时间复杂度：O(n Log n)
辅助空间：O(1)
一个有效的解决方案是使用哈希。我们创建一个哈希表，并将元素及其频率计数存储为键值对。最后，我们遍历哈希表并打印具有最大值的键。

C++

// CPP program to find the most frequent element
// in an array.
#include 
using namespace std;
 
int mostFrequent(int arr[], int n)
{
    // Insert all elements in hash.
    unordered_map hash;
    for (int i = 0; i < n; i++)
        hash[arr[i]]++;
 
    // find the max frequency
    int max_count = 0, res = -1;
    for (auto i : hash) {
        if (max_count < i.second) {
            res = i.first;
            max_count = i.second;
        }
    }
 
    return res;
}
 
// driver program
int main()
{
    int arr[] = { 1, 5, 2, 1, 3, 2, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << mostFrequent(arr, n);
    return 0;
}

Java

//Java program to find the most frequent element
//in an array
import java.util.HashMap;
import java.util.Map;
import java.util.Map.Entry;
 
class GFG {
     
    static int mostFrequent(int arr[], int n)
    {
         
        // Insert all elements in hash
        Map hp =
               new HashMap();
         
        for(int i = 0; i < n; i++)
        {
            int key = arr[i];
            if(hp.containsKey(key))
            {
                int freq = hp.get(key);
                freq++;
                hp.put(key, freq);
            }
            else
            {
                hp.put(key, 1);
            }
        }
         
        // find max frequency.
        int max_count = 0, res = -1;
         
        for(Entry val : hp.entrySet())
        {
            if (max_count < val.getValue())
            {
                res = val.getKey();
                max_count = val.getValue();
            }
        }
         
        return res;
    }
     
    // Driver code
    public static void main (String[] args) {
         
        int arr[] = {1, 5, 2, 1, 3, 2, 1};
        int n = arr.length;
         
        System.out.println(mostFrequent(arr, n));
    }
}
 
// This code is contributed by Akash Singh.

Python3

# Python3 program to find the most
# frequent element in an array.
import math as mt
 
def mostFrequent(arr, n):
 
    # Insert all elements in Hash.
    Hash = dict()
    for i in range(n):
        if arr[i] in Hash.keys():
            Hash[arr[i]] += 1
        else:
            Hash[arr[i]] = 1
 
    # find the max frequency
    max_count = 0
    res = -1
    for i in Hash:
        if (max_count < Hash[i]):
            res = i
            max_count = Hash[i]
         
    return res
 
# Driver Code
arr = [ 1, 5, 2, 1, 3, 2, 1]
n = len(arr)
print(mostFrequent(arr, n))
 
# This code is contributed
# by Mohit kumar 29

C＃

// C# program to find the most
// frequent element in an array
using System;
using System.Collections.Generic;
 
class GFG
{
    static int mostFrequent(int []arr,
                            int n)
    {
        // Insert all elements in hash
        Dictionary hp =
                    new Dictionary();
         
        for (int i = 0; i < n; i++)
        {
            int key = arr[i];
            if(hp.ContainsKey(key))
            {
                int freq = hp[key];
                freq++;
                hp[key] = freq;
            }
            else
                hp.Add(key, 1);
        }
         
        // find max frequency.
        int min_count = 0, res = -1;
         
        foreach (KeyValuePair pair in hp)
        {
            if (min_count < pair.Value)
            {
                res = pair.Key;
                min_count = pair.Value;
            }
        }
        return res;
    }
     
    // Driver code
    static void Main ()
    {
        int []arr = new int[]{1, 5, 2,
                              1, 3, 2, 1};
        int n = arr.Length;
         
        Console.Write(mostFrequent(arr, n));
    }
}
 
// This code is contributed by
// Manish Shaw(manishshaw1)

输出：

1

时间复杂度： O(n)
辅助空间： O(n)