数组中只出现一次的元素

给定一个数组arr ，其中的数字出现了两次或一次。任务是识别在数组中只出现一次的数字。

注意：重复项每次并排出现。可能有几个数字可以同时出现，假设这是一个正确的旋转数组（只是说一个数组可以向右旋转 k 次）。输出中元素的顺序无关紧要。

例子：

Input: arr[] = { 7, 7, 8, 8, 9, 1, 1, 4, 2, 2 }
Output: 9 4

Input: arr[] = {-9, -8, 4, 4, 5, 5, -1}
Output: -9 -8 -1

方法 1：使用排序。

• 对数组进行排序。
• 检查索引 i 处的每个元素（第一个和最后一个元素除外），如果

arr[i] != arr[i-1] && arr [i] != arr[i+1]

• 对于第一个元素，检查是否arr[0] != arr[1] 。
• 对于最后一个元素，检查是否arr[n-1] != arr[n-2] 。

下面是上述方法的实现：

C++

// C++ implementation of above approach
#include 
using namespace std;
 
// Function to find the elements that
// appeared only once in the array
void occurredOnce(int arr[], int n)
{
    // Sort the array
    sort(arr, arr + n);
 
    // Check for first element
    if (arr[0] != arr[1])
        cout << arr[0] << " ";
 
    // Check for all the elements if it is different
    // its adjacent elements
    for (int i = 1; i < n - 1; i++)
        if (arr[i] != arr[i + 1] && arr[i] != arr[i - 1])
            cout << arr[i] << " ";
 
    // Check for the last element
    if (arr[n - 2] != arr[n - 1])
        cout << arr[n - 1] << " ";
}
 
// Driver code
int main()
{
 
    int arr[] = { 7, 7, 8, 8, 9, 1, 1, 4, 2, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    occurredOnce(arr, n);
 
    return 0;
}

Java

// Java implementation
// of above approach
import java.util.*;
 
class GFG
{
 
// Function to find the elements that
// appeared only once in the array
static void occurredOnce(int arr[], int n)
{
    // Sort the array
    Arrays.sort(arr);
 
    // Check for first element
    if (arr[0] != arr[1])
        System.out.println(arr[0] + " ");
 
    // Check for all the elements
    // if it is different
    // its adjacent elements
    for (int i = 1; i < n - 1; i++)
        if (arr[i] != arr[i + 1] &&
            arr[i] != arr[i - 1])
            System.out.print(arr[i] + " ");
 
    // Check for the last element
    if (arr[n - 2] != arr[n - 1])
        System.out.print(arr[n - 1] + " ");
}
 
// Driver code
public static void main(String args[])
{
    int arr[] = {7, 7, 8, 8, 9,
                 1, 1, 4, 2, 2};
    int n = arr.length;
 
    occurredOnce(arr, n);
}
}
 
// This code is contributed
// by Arnab Kundu

Python3

# Python 3 implementation
# of above approach
 
# Function to find the elements
# that appeared only once in
# the array
def occurredOnce(arr, n):
     
    # Sort the array
    arr.sort()
 
    # Check for first element
    if arr[0] != arr[1]:
        print(arr[0], end = " ")
 
    # Check for all the elements
    # if it is different its
    # adjacent elements
    for i in range(1, n - 1):
        if (arr[i] != arr[i + 1] and
            arr[i] != arr[i - 1]):
            print( arr[i], end = " ")
 
    # Check for the last element
    if arr[n - 2] != arr[n - 1]:
        print(arr[n - 1], end = " ")
 
# Driver code
if __name__ == "__main__":
 
    arr = [ 7, 7, 8, 8, 9,
            1, 1, 4, 2, 2 ]
    n = len(arr)
    occurredOnce(arr, n)
 
# This code is contributed
# by ChitraNayal

C#

// C# implementation
// of above approach
using System;
 
class GFG
{
 
// Function to find the elements that
// appeared only once in the array
static void occurredOnce(int[] arr, int n)
{
    // Sort the array
    Array.Sort(arr);
 
    // Check for first element
    if (arr[0] != arr[1])
        Console.Write(arr[0] + " ");
 
    // Check for all the elements
    // if it is different
    // its adjacent elements
    for (int i = 1; i < n - 1; i++)
        if (arr[i] != arr[i + 1] &&
            arr[i] != arr[i - 1])
            Console.Write(arr[i] + " ");
 
    // Check for the last element
    if (arr[n - 2] != arr[n - 1])
        Console.Write(arr[n - 1] + " ");
}
 
// Driver code
public static void Main()
{
    int[] arr = {7, 7, 8, 8, 9,
                1, 1, 4, 2, 2};
    int n = arr.Length;
 
    occurredOnce(arr, n);
}
}
 
// This code is contributed
// by ChitraNayal

Javascript

 
// Javascript implementation
// of above approach
 
// Function to find the elements that
// appeared only once in the array
function occurredOnce(arr,n)
{
    // Sort the array
    arr.sort(function(a,b){return a-b;});
  
    // Check for first element
    if (arr[0] != arr[1])
        document.write(arr[0] + " ");
  
    // Check for all the elements
    // if it is different
    // its adjacent elements
    for (let i = 1; i < n - 1; i++)
        if (arr[i] != arr[i + 1] &&
            arr[i] != arr[i - 1])
            document.write(arr[i] + " ");
  
    // Check for the last element
    if (arr[n - 2] != arr[n - 1])
        document.write(arr[n - 1] + " ");
}
 
// Driver code
let arr=[7, 7, 8, 8, 9,
                 1, 1, 4, 2, 2];
let n = arr.length;
occurredOnce(arr, n);
 
 
// This code is contributed by rag2127
 

输出

4 9 

时间复杂度： O(Nlogn)
空间复杂度： O(1)

方法 2：（使用散列） ：在 C++ 中，unordered_map 可用于散列。

• 遍历数组。
• 将每个元素及其出现情况存储在 unordered_map 中。
• 遍历 unordered_map 并打印出现次数为 1 的所有元素。

下面是上述方法的实现：

C++

// C++ implementation to find elements
// that appeared only once
#include 
using namespace std;
 
// Function to find the elements that
// appeared only once in the array
void occurredOnce(int arr[], int n)
{
    unordered_map mp;
 
    // Store all the elements in the map with
    // their occurrence
    for (int i = 0; i < n; i++)
        mp[arr[i]]++;
 
    // Traverse the map and print all the
    // elements with occurrence 1
    for (auto it = mp.begin(); it != mp.end(); it++)
        if (it->second == 1)
            cout << it->first << " ";
}
 
// Driver code
int main()
{
 
    int arr[] = { 7, 7, 8, 8, 9, 1, 1, 4, 2, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    occurredOnce(arr, n);
 
    return 0;
}

Java

// Java implementation to find elements
// that appeared only once
import java.util.*;
import java.io.*;
 
class GFG
{
    // Function to find the elements that
    // appeared only once in the array
    static void occurredOnce(int[] arr, int n)
    {
            HashMap mp = new HashMap<>();
             
            // Store all the elements in the map with
            // their occurrence
            for (int i = 0; i < n; i++)
            {
                if (mp.containsKey(arr[i]))
                    mp.put(arr[i], 1 + mp.get(arr[i]));
                else
                    mp.put(arr[i], 1);
            }
 
            // Traverse the map and print all the
            // elements with occurrence 1
            for (Map.Entry entry : mp.entrySet())
            {
                if (Integer.parseInt(String.valueOf(entry.getValue())) == 1)
                    System.out.print(entry.getKey() + " ");
            }
    }
 
    // Driver code
    public static void main(String args[])
    {
            int[] arr = { 7, 7, 8, 8, 9, 1, 1, 4, 2, 2 };
            int n = arr.length;
 
            occurredOnce(arr, n);
    }
}
 
// This code is contributed by rachana soma

蟒蛇3

# Python3 implementation to find elements
# that appeared only once
import math as mt
 
# Function to find the elements that
# appeared only once in the array
def occurredOnce(arr, n):
 
    mp = dict()
 
    # Store all the elements in the
    # map with their occurrence
    for i in range(n):
        if arr[i] in mp.keys():
            mp[arr[i]] += 1
        else:
            mp[arr[i]] = 1
 
    # Traverse the map and print all
    # the elements with occurrence 1
    for it in mp:
        if mp[it] == 1:
            print(it, end = " ")
 
# Driver code
arr = [7, 7, 8, 8, 9, 1, 1, 4, 2, 2]
n = len(arr)
 
occurredOnce(arr, n)
 
# This code is contributed by
# Mohit Kumar 29

C＃

// C# implementation to find elements
// that appeared only once
using System;
using System.Collections.Generic;
class GFG
{
    // Function to find the elements that
    // appeared only once in the array
    static void occurredOnce(int[] arr, int n)
    {
      Dictionary mp = new Dictionary();
 
      // Store all the elements in the map with
      // their occurrence
      for (int i = 0; i < n; i++)
      {
        if (mp.ContainsKey(arr[i]))
          mp[arr[i]] = 1 + mp[arr[i]];
        else
          mp.Add(arr[i], 1);
      }
 
      // Traverse the map and print all the
      // elements with occurrence 1
      foreach(KeyValuePair entry in mp)
      {
        if (Int32.Parse(String.Join("", entry.Value)) == 1)
          Console.Write(entry.Key + " ");
      }
    }
 
    // Driver code
    public static void Main(String []args)
    {
      int[] arr = { 7, 7, 8, 8, 9, 1, 1, 4, 2, 2 };
      int n = arr.Length;
      occurredOnce(arr, n);
    }
}
 
// This code is contributed by shikhasingrajput

Javascript

 
// Javascript implementation to find elements
// that appeared only once
     
// Function to find the elements that
// appeared only once in the array
function occurredOnce(arr, n)
{
    let mp = new Map();
     
    // Store all the elements in the map
    // with their occurrence
    for(let i = 0; i < n; i++)
    {
        if (mp.has(arr[i]))
            mp.set(arr[i], 1 + mp.get(arr[i]));
        else
            mp.set(arr[i], 1);
    }
 
    // Traverse the map and print all the
    // elements with occurrence 1
    for(let [key, value] of mp.entries())
    {
        if (value == 1)
            document.write(key + " ");
    }
}
 
// Driver code
let arr = [ 7, 7, 8, 8, 9,
            1, 1, 4, 2, 2 ];
let n = arr.length;
 
occurredOnce(arr, n);
 
// This code is contributed by avanitrachhadiya2155
 

输出

4 9 

时间复杂度： O(N)
空间复杂度： O(N)

方法 3：使用给定的假设。
假设数组可以随时旋转，并且每次都会并排出现重复项。因此，旋转后，第一个和最后一个元素将并排出现。

• 检查第一个和最后一个元素是否相等。如果是，则开始遍历它们之间的元素。
• 检查当前元素是否等于前一个索引中的元素。如果是，请检查下一个元素是否相同。
• 如果没有，则打印当前元素。

C++

// C++ implementation to find elements
// that appeared only once
#include 
using namespace std;
 
// Function to find the elements that
// appeared only once in the array
void occurredOnce(int arr[], int n)
{
    int i = 1, len = n;
 
    // Check if the first and last element is equal
    // If yes, remove those elements
    if (arr[0] == arr[len - 1]) {
        i = 2;
        len--;
    }
 
    // Start traversing the remaining elements
    for (; i < n; i++)
 
        // Check if current element is equal to
        // the element at immediate previous index
        // If yes, check the same for next element
        if (arr[i] == arr[i - 1])
            i++;
 
        // Else print the current element
        else
            cout << arr[i - 1] << " ";
 
    // Check for the last element
    if (arr[n - 1] != arr[0] && arr[n - 1] != arr[n - 2])
        cout << arr[n - 1];
}
 
// Driver code
int main()
{
 
    int arr[] = { 7, 7, 8, 8, 9, 1, 1, 4, 2, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    occurredOnce(arr, n);
 
    return 0;
}

Java

// Java implementation to find
// elements that appeared only once
class GFG
{
// Function to find the elements that
// appeared only once in the array
static void occurredOnce(int arr[], int n)
{
    int i = 1, len = n;
 
    // Check if the first and last
    // element is equal. If yes,
    // remove those elements
    if (arr[0] == arr[len - 1])
    {
        i = 2;
        len--;
    }
 
    // Start traversing the
    // remaining elements
    for (; i < n; i++)
 
        // Check if current element is
        // equal to the element at
        // immediate previous index
        // If yes, check the same
        // for next element
        if (arr[i] == arr[i - 1])
            i++;
 
        // Else print the current element
        else
            System.out.print(arr[i - 1] + " ");
 
    // Check for the last element
    if (arr[n - 1] != arr[0] &&
        arr[n - 1] != arr[n - 2])
        System.out.print(arr[n - 1]);
}
 
// Driver code
public static void main(String args[])
{
    int arr[] = {7, 7, 8, 8, 9,
                 1, 1, 4, 2, 2};
    int n = arr.length;
 
    occurredOnce(arr, n);
}
}
 
// This code is contributed
// by Arnab Kundu

蟒蛇3

# Python 3 implementation to find
# elements that appeared only once
 
# Function to find the elements that
# appeared only once in the array
def occurredOnce(arr, n):
    i = 1
    len = n
 
    # Check if the first and
    # last element is equal
    # If yes, remove those elements
    if arr[0] == arr[len - 1]:
        i = 2
        len -= 1
 
    # Start traversing the
    # remaining elements
    while i < n:
 
        # Check if current element is
        # equal to the element at
        # immediate previous index
        # If yes, check the same for
        # next element
        if arr[i] == arr[i - 1]:
            i += 1
 
        # Else print the current element
        else:
            print(arr[i - 1], end = " ")
             
        i += 1
 
    # Check for the last element
    if (arr[n - 1] != arr[0] and
        arr[n - 1] != arr[n - 2]):
        print(arr[n - 1])
 
# Driver code
if __name__ == "__main__":
    arr = [ 7, 7, 8, 8, 9, 1, 1, 4, 2, 2 ]
    n = len(arr)
 
    occurredOnce(arr, n)
 
# This code is contributed
# by ChitraNayal

C＃

// C# implementation to find
// elements that appeared only once
using System;
 
class GFG
{
// Function to find the elements that
// appeared only once in the array
static void occurredOnce(int[] arr, int n)
{
    int i = 1, len = n;
 
    // Check if the first and last
    // element is equal. If yes,
    // remove those elements
    if (arr[0] == arr[len - 1])
    {
        i = 2;
        len--;
    }
 
    // Start traversing the
    // remaining elements
    for (; i < n; i++)
 
        // Check if current element is
        // equal to the element at
        // immediate previous index
        // If yes, check the same
        // for next element
        if (arr[i] == arr[i - 1])
            i++;
 
        // Else print the current element
        else
            Console.Write(arr[i - 1] + " ");
 
    // Check for the last element
    if (arr[n - 1] != arr[0] &&
        arr[n - 1] != arr[n - 2])
        Console.Write(arr[n - 1]);
}
 
// Driver code
public static void Main()
{
    int[] arr = {7, 7, 8, 8, 9,
                1, 1, 4, 2, 2};
    int n = arr.Length;
 
    occurredOnce(arr, n);
}
}
 
// This code is contributed
// by ChitraNayal

Javascript

 
// Javascript implementation to find
// elements that appeared only once
 
// Function to find the elements that
// appeared only once in the array
function occurredOnce(arr, n)
{
    var i = 1, len = n;
 
    // Check if the first and last
    // element is equal. If yes,
    // remove those elements
    if (arr[0] == arr[len - 1])
    {
        i = 2;
        len--;
    }
 
    // Start traversing the
    // remaining elements
    for(; i < n; i++)
 
        // Check if current element is
        // equal to the element at
        // immediate previous index
        // If yes, check the same
        // for next element
        if (arr[i] == arr[i - 1])
            i++;
 
        // Else print the current element
        else
            document.write(arr[i - 1] + " ");
 
    // Check for the last element
    if (arr[n - 1] != arr[0] &&
        arr[n - 1] != arr[n - 2])
        document.write(arr[n - 1]);
}
 
// Driver code
var arr = [ 7, 7, 8, 8, 9,
            1, 1, 4, 2, 2 ];
var n = arr.length;
 
occurredOnce(arr, n);
 
// This code is contributed by Ankita saini
 

输出

9 4 

时间复杂度： O(N)
空间复杂度： O(1)

方法#4：使用内置的Python函数：

• 使用 Counter函数计算每个元素的频率
• 遍历频率数组并打印出现次数为 1 的所有元素。

下面是实现

蟒蛇3

# Python3 implementation to find elements
# that appeared only once
from collections import Counter
 
# Function to find the elements that
# appeared only once in the array
def occurredOnce(arr, n):
 
    #counting frequency of every element using Counter
    mp=Counter(arr)
    # Traverse the map and print all
    # the elements with occurrence 1
    for it in mp:
        if mp[it] == 1:
            print(it, end = " ")
 
# Driver code
arr = [7, 7, 8, 8, 9, 1, 1, 4, 2, 2]
n = len(arr)
 
occurredOnce(arr, n)
 
# This code is contributed by vikkycirus

输出

9 4 

时间复杂度： O(n)

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