查找出现次数的奇数

给定正整数数组。除了一个数字出现奇数次以外，所有数字均出现偶数次。在O(n)时间和恒定空间中找到数字。
例子 ：

Input : arr = {1, 2, 3, 2, 3, 1, 3}
Output : 3

Input : arr = {5, 7, 2, 7, 5, 2, 5}
Output : 5

一个简单的解决方案是运行两个嵌套循环。外循环一个接一个地拾取所有元素，内循环计算外循环拾取的元素出现的次数。该解决方案的时间复杂度为O(n ² )。
下面是蛮力方法的实现：

C++

// C++ program to find the element
// occurring odd number of times
#include
using namespace std;
 
// Function to find the element
// occurring odd number of times
int getOddOccurrence(int arr[], int arr_size)
{
    for (int i = 0; i < arr_size; i++) {
         
        int count = 0;
         
        for (int j = 0; j < arr_size; j++)
        {
            if (arr[i] == arr[j])
                count++;
        }
        if (count % 2 != 0)
            return arr[i];
    }
    return -1;
}
 
// driver code
int main()
    {
        int arr[] = { 2, 3, 5, 4, 5, 2,
                      4, 3, 5, 2, 4, 4, 2 };
        int n = sizeof(arr) / sizeof(arr[0]);
 
        // Function calling
        cout << getOddOccurrence(arr, n);
 
        return 0;
    }

Java

// Java program to find the element occurring
// odd number of times
class OddOccurrence {
     
    // function to find the element occurring odd
    // number of times
    static int getOddOccurrence(int arr[], int arr_size)
    {
        int i;
        for (i = 0; i < arr_size; i++) {
            int count = 0;
            for (int j = 0; j < arr_size; j++) {
                if (arr[i] == arr[j])
                    count++;
            }
            if (count % 2 != 0)
                return arr[i];
        }
        return -1;
    }
     
    // driver code
    public static void main(String[] args)
    {
        int arr[] = new int[]{ 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 };
        int n = arr.length;
        System.out.println(getOddOccurrence(arr, n));
    }
}
// This code has been contributed by Kamal Rawal

Python3

# Python program to find the element occurring
# odd number of times
     
# function to find the element occurring odd
# number of times
def getOddOccurrence(arr, arr_size):
     
    for i in range(0,arr_size):
        count = 0
        for j in range(0, arr_size):
            if arr[i] == arr[j]:
                count+=1
             
        if (count % 2 != 0):
            return arr[i]
         
    return -1
     
     
# driver code
arr = [2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 ]
n = len(arr)
print(getOddOccurrence(arr, n))
 
# This code has been contributed by
# Smitha Dinesh Semwal

C#

// C# program to find the element
// occurring odd number of times
using System;
 
class GFG
{
    // Function to find the element
    // occurring odd number of times
    static int getOddOccurrence(int []arr, int arr_size)
    {
        for (int i = 0; i < arr_size; i++) {
            int count = 0;
             
            for (int j = 0; j < arr_size; j++) {
                if (arr[i] == arr[j])
                    count++;
            }
            if (count % 2 != 0)
                return arr[i];
        }
        return -1;
    }
     
    // Driver code
    public static void Main()
    {
        int []arr = { 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 };
        int n = arr.Length;
        Console.Write(getOddOccurrence(arr, n));
    }
}
 
// This code is contributed by Sam007

Javascript

 
// Javascript program to find the element
// occurring odd number of times
 
// Function to find the element
// occurring odd number of times
function getOddOccurrence(arr, arr_size)
{
    for (let i = 0; i < arr_size; i++) {
         
        let count = 0;
         
        for (let j = 0; j < arr_size; j++)
        {
            if (arr[i] == arr[j])
                count++;
        }
        if (count % 2 != 0)
            return arr[i];
    }
    return -1;
}
 
// driver code
 
        let arr = [ 2, 3, 5, 4, 5, 2,
                    4, 3, 5, 2, 4, 4, 2 ];
        let n = arr.length;
 
        // Function calling
        document.write(getOddOccurrence(arr, n));
 
// This code is contributed by Mayank Tyagi
 

输出 ：

5

更好的解决方案是使用哈希。使用数组元素作为键，并将其计数作为值。创建一个空的哈希表。一对一遍历给定的数组元素并存储计数。该解决方案的时间复杂度为O(n)。但是它需要额外的空间来进行哈希处理。
程序 ：

C++

// C++ program to find the element 
// occurring odd number of times
#include 
using namespace std;
 
// function to find the element
// occurring odd number of times
int getOddOccurrence(int arr[],int size)
{
     
    // Defining HashMap in C++
    unordered_map hash;
 
    // Putting all elements into the HashMap
    for(int i = 0; i < size; i++)
    {
        hash[arr[i]]++;
    }
    // Iterate through HashMap to check an element
    // occurring odd number of times and return it
    for(auto i : hash)
    {
        if(i.second % 2 != 0)
        {
            return i.first;
        }
    }
    return -1;
}
 
// Driver code
int main()
{
    int arr[] = { 2, 3, 5, 4, 5, 2, 4,
                    3, 5, 2, 4, 4, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
     
    // Function calling
    cout << getOddOccurrence(arr, n);
 
    return 0;
}
 
// This code is contributed by codeMan_d.

Java

// Java program to find the element occurring odd
// number of times
import java.io.*;
import java.util.HashMap;
 
class OddOccurrence
{
    // function to find the element occurring odd
    // number of times
    static int getOddOccurrence(int arr[], int n)
    {
        HashMap hmap = new HashMap<>();
         
        // Putting all elements into the HashMap
        for(int i = 0; i < n; i++)
        {
            if(hmap.containsKey(arr[i]))
            {
                int val = hmap.get(arr[i]);
                         
                // If array element is already present then
                // increase the count of that element.
                hmap.put(arr[i], val + 1);
            }
            else
                 
                // if array element is not present then put
                // element into the HashMap and initialize
                // the count to one.
                hmap.put(arr[i], 1);
        }
 
        // Checking for odd occurrence of each element present
        // in the HashMap
        for(Integer a:hmap.keySet())
        {
            if(hmap.get(a) % 2 != 0)
                return a;
        }
        return -1;
    }
         
    // driver code   
    public static void main(String[] args)
    {
        int arr[] = new int[]{2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2};
        int n = arr.length;
        System.out.println(getOddOccurrence(arr, n));
    }
}
// This code is contributed by Kamal Rawal

Python3

# Python3 program to find the element 
# occurring odd number of times
  
# function to find the element
# occurring odd number of times
def getOddOccurrence(arr,size):
      
    # Defining HashMap in C++
    Hash=dict()
  
    # Putting all elements into the HashMap
    for i in range(size):
        Hash[arr[i]]=Hash.get(arr[i],0) + 1;
     
    # Iterate through HashMap to check an element
    # occurring odd number of times and return it
    for i in Hash:
 
        if(Hash[i]% 2 != 0):
            return i
    return -1
 
  
# Driver code
arr=[2, 3, 5, 4, 5, 2, 4,3, 5, 2, 4, 4, 2]
n = len(arr)
  
# Function calling
print(getOddOccurrence(arr, n))
 
# This code is contributed by mohit kumar

C＃

// C# program to find the element occurring odd
// number of times
using System;
using System.Collections.Generic;
 
public class OddOccurrence
{
    // function to find the element occurring odd
    // number of times
    static int getOddOccurrence(int []arr, int n)
    {
        Dictionary hmap = new Dictionary();
         
        // Putting all elements into the HashMap
        for(int i = 0; i < n; i++)
        {
            if(hmap.ContainsKey(arr[i]))
            {
                int val = hmap[arr[i]];
                         
                // If array element is already present then
                // increase the count of that element.
                hmap.Remove(arr[i]);
                hmap.Add(arr[i], val + 1);
            }
            else
                 
                // if array element is not present then put
                // element into the HashMap and initialize
                // the count to one.
                hmap.Add(arr[i], 1);
        }
 
        // Checking for odd occurrence of each element present
        // in the HashMap
        foreach(KeyValuePair entry in hmap)
        {
            if(entry.Value % 2 != 0)
            {
                return entry.Key;
            }
        }
        return -1;
    }
         
    // Driver code
    public static void Main(String[] args)
    {
        int []arr = new int[]{2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2};
        int n = arr.Length;
        Console.WriteLine(getOddOccurrence(arr, n));
    }
}
 
// This code is contributed by Princi Singh

Java脚本

// Javascript program to find the element occurring odd
// number of times
     
    // function to find the element occurring odd
    // number of times
    function getOddOccurrence(arr,n)
    {
        let hmap = new Map();
          
        // Putting all elements into the HashMap
        for(let i = 0; i < n; i++)
        {
            if(hmap.has(arr[i]))
            {
                let val = hmap.get(arr[i]);
                // If array element is already present then
                // increase the count of that element.
                hmap.set(arr[i], val + 1);
            }
            else
            {    
                // if array element is not present then put
                // element into the HashMap and initialize
                // the count to one.
                hmap.set(arr[i], 1);
            }
        }
         
  
        // Checking for odd occurrence of each element present
        // in the HashMap
        for(let [key, value] of hmap.entries())
        {
            //document.write(hmap[a]+"<br>")
            if(hmap.get(key) % 2 != 0)
                return key;
        }
        return -1;
    }
     
    // driver code  
    let arr=[2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2];
    let n = arr.length;
    document.write(getOddOccurrence(arr, n));
     
     
    // This code is contributed by unknown2108

输出 ：

5

最佳解决方案是对所有元素进行按位XOR。所有元素的XOR给我们奇怪的发生元素。请注意，如果两个元素相同，则两个元素的XOR为0，且数字x与0的XOR为x。
下面是上述方法的实现。

C++

// C++ program to find the element
// occurring odd number of times
#include 
using namespace std;
 
// Function to find element occurring
// odd number of times
int getOddOccurrence(int ar[], int ar_size)
{
    int res = 0;
    for (int i = 0; i < ar_size; i++)    
        res = res ^ ar[i];
     
    return res;
}
 
/* Driver function to test above function */
int main()
{
    int ar[] = {2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2};
    int n = sizeof(ar)/sizeof(ar[0]);
     
    // Function calling
    cout << getOddOccurrence(ar, n);
     
    return 0;
}

C

// C program to find the element
// occurring odd number of times
#include 
 
// Function to find element occurring
// odd number of times
int getOddOccurrence(int ar[], int ar_size)
{
    int res = 0;
    for (int i = 0; i < ar_size; i++)    
        res = res ^ ar[i];
     
    return res;
}
 
/* Driver function to test above function */
int main()
{
    int ar[] = {2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2};
    int n = sizeof(ar) / sizeof(ar[0]);
     
    // Function calling
    printf("%d", getOddOccurrence(ar, n));
    return 0;
}

Java

//Java program to find the element occurring odd number of times
 
class OddOccurance
{
    int getOddOccurrence(int ar[], int ar_size)
    {
        int i;
        int res = 0;
        for (i = 0; i < ar_size; i++)
        {
            res = res ^ ar[i];
        }
        return res;
    }
 
    public static void main(String[] args)
    {
        OddOccurance occur = new OddOccurance();
        int ar[] = new int[]{2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2};
        int n = ar.length;
        System.out.println(occur.getOddOccurrence(ar, n));
    }
}
// This code has been contributed by Mayank Jaiswal

Python3

# Python program to find the element occurring odd number of times
 
def getOddOccurrence(arr):
 
    # Initialize result
    res = 0
     
    # Traverse the array
    for element in arr:
        # XOR with the result
        res = res ^ element
 
    return res
 
# Test array
arr = [ 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2]
 
print("%d" % getOddOccurrence(arr))

C＃

// C# program to find the element
// occurring odd number of times
using System;
 
class GFG
{
    // Function to find the element
    // occurring odd number of times
    static int getOddOccurrence(int []arr, int arr_size)
    {
        int res = 0;
        for (int i = 0; i < arr_size; i++)
        {
            res = res ^ arr[i];
        }
        return res;
    }
     
    // Driver code
    public static void Main()
    {
        int []arr = { 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 };
        int n = arr.Length;
        Console.Write(getOddOccurrence(arr, n));
    }
}
 
// This code is contributed by Sam007

Java脚本

 
// JavaScript program to find the element
// occurring odd number of times
 
// Function to find the element
// occurring odd number of times
function getOddOccurrence( ar, ar_size)
{
    let res = 0;
     
    for (let i = 0; i < ar_size; i++)    
        res = res ^ ar[i];
     
    return res;
}
 
    // driver code
 
    let arr = [ 2, 3, 5, 4, 5, 2, 4,
                3, 5, 2, 4, 4, 2 ];
    let n = arr.length;
 
    // Function calling
    document.write(getOddOccurrence(arr, n));
 
 

输出 ：

5

时间复杂度： O(n)

方法3：使用内置的Python函数：

• 使用计数器函数计算每个元素的频率
• 遍历频率字典
• 检查哪个元素具有奇数频率。
• 打印该元素并打破循环

下面是实现：

Python3

# importing counter from colleections
from collections import Counter
 
# Python3 implementation to find
# odd frequeency element
def oddElement(arr, n):
 
    # Calculating freequencies usinf Counter
    count_map = Counter(arr)
 
    for i in range(0, n):
 
        # If count of element is odd we return
        if (count_map[arr[i]] % 2 != 0):
            return arr[i]
 
 
# Driver Code
if __name__ == "__main__":
 
    arr = [1, 1, 3, 3, 5, 6, 6]
    n = len(arr)
    print(oddElement(arr, n))
 
# This code is contributed by vikkycirus

输出：

5

时间复杂度： O(N)