查找1到n-1之间的唯一重复元素

给我们一个大小为n的数组arr []。编号从1到(n-1)以随机顺序排列。该数组只有一个重复元素。我们需要找到重复元素。

例子 ：

Input  : a[] = {1, 3, 2, 3, 4}
Output : 3

Input  : a[] = {1, 5, 1, 2, 3, 4}
Output : 1

方法1(简单)我们使用两个嵌套循环。外循环遍历所有元素，内循环检查外循环拾取的元素是否出现在其他任何地方。

时间复杂度： O(n * n)

方法2(使用求和公式)：我们知道前n-1个自然数的总和为(n – 1)* n / 2。我们计算数组元素的总和，并从中减去自然数总和，以找到唯一的缺失元素。

C++

// CPP program to find the only repeating
// element in an array where elements are
// from 1 to n-1.
#include 
using namespace std;
  
int findRepeating(int arr[], int n)
{
   // Find array sum and subtract sum
   // first n-1 natural numbers from it
   // to find the result.
   return accumulate(arr , arr+n , 0) - 
                   ((n - 1) * n/2);
}
  
// driver code
int main()
{   
    int arr[] = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << findRepeating(arr, n);
    return 0;
}

Java

// Java program to find the only repeating 
// element in an array where elements are 
// from 1 to n-1.
import java.io.*;
import java.util.*; 
  
class GFG 
{
  
    static int findRepeating(int[] arr, int n)
    {
        // Find array sum and subtract sum 
        // first n-1 natural numbers from it 
        // to find the result.
  
        int sum = 0;
        for (int i = 0; i < n; i++)
            sum += arr[i];
        return sum - (((n - 1) * n) / 2); 
    }
  
    // Driver code
    public static void main(String args[]) 
    {
        int[] arr = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };
        int n = arr.length;
        System.out.println(findRepeating(arr, n));
    }
}
  
// This code is contributed by rachana soma

Python3

# Python3 program to find the only 
# repeating element in an array where 
# elements are from 1 to n-1.
  
def findRepeating(arr, n):
      
    # Find array sum and subtract sum
    # first n-1 natural numbers from it
    # to find the result.
    return sum(arr) -(((n - 1) * n) // 2)
  
# Driver Code
arr = [9, 8, 2, 6, 1, 8, 5, 3, 4, 7]
n = len(arr)
print(findRepeating(arr, n))
  
# This code is contributed 
# by mohit kumar

C#

// C# program to find the only repeating 
// element in an array where elements are 
// from 1 to n-1.
using System; 
  
class GFG 
{
  
    static int findRepeating(int[] arr, int n)
    {
        // Find array sum and subtract sum 
        // first n-1 natural numbers from it 
        // to find the result.
  
        int sum = 0;
        for (int i = 0; i < n; i++)
            sum += arr[i];
        return sum - (((n - 1) * n) / 2); 
    }
  
    // Driver code
    public static void Main(String []args) 
    {
        int[] arr = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };
        int n = arr.Length;
        Console.WriteLine(findRepeating(arr, n));
    }
}
  
/* This code contributed by PrinciRaj1992 */

---

---

输出 ：

8

时间复杂度： O(n)
辅助空间： O(1)
导致大型阵列溢出。

方法3(使用哈希)：使用哈希表存储访问的元素。如果看到的元素再次出现，我们将其返回。

C++

// CPP program to find the only repeating
// element in an array where elements are
// from 1 to n-1.
#include 
using namespace std;
  
int findRepeating(int arr[], int n)
{
   unordered_set s;
   for (int i=0; i

Java

import java.util.*;
// Java program to find the only repeating 
// element in an array where elements are 
// from 1 to n-1.
  
class GFG 
{
  
static int findRepeating(int arr[], int n) 
{ 
    HashSet s = new HashSet();
    for (int i = 0; i < n; i++) 
    { 
        if (s.contains(arr[i])) 
            return arr[i]; 
        s.add(arr[i]); 
    } 
      
    // If input is correct, we should 
    // never reach here 
    return -1; 
} 
  
// Driver code 
public static void main(String[] args) 
{
    int arr[] = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 }; 
    int n = arr.length;
    System.out.println(findRepeating(arr, n));;
}
}
  
// This code is contributed by Rajput-Ji

Python3

# Python3 program to find the only 
# repeating element in an array 
# where elements are from 1 to n-1. 
def findRepeating(arr, n):
    s = set()
    for i in range(n):
        if arr[i] in s:
            return arr[i]
        s.add(arr[i])
      
    # If input is correct, we should 
    # never reach here 
    rteurn -1
  
# Driver code
arr = [9, 8, 2, 6, 1, 8, 5, 3]
n = len(arr)
print(findRepeating(arr, n))
  
# This code is contributed 
# by Shrikant13

C＃

// C# program to find the only repeating 
// element in an array where elements are 
// from 1 to n-1.
using System;
using System.Collections.Generic;
  
class GFG 
{
  
static int findRepeating(int []arr, int n) 
{ 
    HashSet s = new HashSet();
    for (int i = 0; i < n; i++) 
    { 
        if (s.Contains(arr[i])) 
            return arr[i]; 
        s.Add(arr[i]); 
    } 
      
    // If input is correct, we should 
    // never reach here 
    return -1; 
} 
  
// Driver code 
public static void Main(String[] args) 
{
    int []arr = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 }; 
    int n = arr.Length;
    Console.WriteLine(findRepeating(arr, n));;
}
}
  
// This code has been contributed by 29AjayKumar

输出 ：

8

时间复杂度： O(n)
辅助空间： O(n)

方法4(使用XOR)：该想法基于x ^ x = 0且x ^ y = y ^ x的事实。
1)计算从1到n-1的元素的XOR。
2)计算数组元素的XOR。
3)以上两项的异或将是我们的结果。

C++

// CPP program to find the only repeating
// element in an array where elements are
// from 1 to n-1.
#include 
using namespace std;
  
int findRepeating(int arr[], int n)
{
  
   // res is going to store value of
   // 1 ^ 2 ^ 3 .. ^ (n-1) ^ arr[0] ^ 
   // arr[1] ^ .... arr[n-1]
   int res = 0;
   for (int i=0; i

Java

// Java program to find the only repeating
// element in an array where elements are
// from 1 to n-1.
class GFG
{
      
    static int findRepeating(int arr[], int n)
    {
      
        // res is going to store value of
        // 1 ^ 2 ^ 3 .. ^ (n-1) ^ arr[0] ^ 
        // arr[1] ^ .... arr[n-1]
        int res = 0;
        for (int i = 0; i < n - 1; i++)
            res = res ^ (i + 1) ^ arr[i];
        res = res ^ arr[n - 1];
              
        return res;
    }
      
    // Driver code
    public static void main(String[] args)
    { 
        int arr[] = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };
        int n = arr.length;
        System.out.println(findRepeating(arr, n));
    }
}
  
// This code is contributed by 
// Smitha Dinesh Semwal.

Python3

# Python3 program to find the only 
# repeating element in an array where 
# elements are from 1 to n-1.
  
def findRepeating(arr, n):
      
    # res is going to store value of
    # 1 ^ 2 ^ 3 .. ^ (n-1) ^ arr[0] ^ 
    # arr[1] ^ .... arr[n-1]
    res = 0
    for i in range(0, n-1):
        res = res ^ (i+1) ^ arr[i]
    res = res ^ arr[n-1]
          
    return res
      
# Driver code
arr = [9, 8, 2, 6, 1, 8, 5, 3, 4, 7] 
n = len(arr) 
print(findRepeating(arr, n))
  
# This code is contributed by Smitha Dinesh Semwal.

C＃

// C# program to find the 
// only repeating element 
// in an array where elements 
// are from 1 to n-1.
using System;
  
class GFG
{
    static int findRepeating(int []arr, 
                             int n)
    {
      
        // res is going to store 
        // value of 1 ^ 2 ^ 3 .. 
        // ^ (n-1) ^ arr[0] ^ 
        // arr[1] ^ .... arr[n-1]
        int res = 0;
        for (int i = 0; i < n - 1; i++)
            res = res ^ (i + 1) ^ arr[i];
        res = res ^ arr[n - 1];
              
        return res;
    }
      
    // Driver code
    public static void Main()
    { 
        int []arr = { 9, 8, 2, 6, 1, 
                      8, 5, 3, 4, 7 };
        int n = arr.Length;
        Console.Write(findRepeating(arr, n));
    }
}
  
// This code is contributed
// by Smitha Dinesh Semwal.

输出：

8

时间复杂度：O(n)
辅助空间：O(1)

方法5：使用索引。
1.遍历数组。
2.对于每个索引访问a [index]，如果为正，则更改a [index]索引处元素的符号，否则打印该元素。

C++

// CPP program to find the only 
// repeating element in an array 
// where elements are from 1 to n-1.
#include 
using namespace std;
  
// Function to find repeted element
int findRepeating(int arr[], int n)
{
    int missingElement = 0;
  
    // indexing based
    for (int i = 0; i < n; i++){
  
        int element = arr[abs(arr[i])];
  
        if(element < 0){
            missingElement = arr[i];
            break;
        }
      
    arr[abs(arr[i])] = -arr[abs(arr[i])];
}
  
return abs(missingElement);
  
}
  
// driver code
int main()
{
    int arr[] = { 5, 4, 3, 9, 8,
                  9, 1, 6, 2, 5};
  
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << findRepeating(arr, n);
  
    return 0;
}

Java

// Java program to find the only 
// repeating element in an array 
// where elements are from 1 to n-1.
import java.lang.Math.*;
  
class GFG
{
    // Function to find repeted element
    static int findRepeating(int arr[], int n)
    {
        int missingElement = 0;
      
        // indexing based
        for (int i = 0; i < n; i++){
      
            int element = arr[Math.abs(arr[i])];
      
            if(element < 0){
                missingElement = arr[i];
                break;
            }
          
        arr[Math.abs(arr[i])] = -arr[Math.abs(arr[i])];
    }
      
    return Math.abs(missingElement);
      
    }
      
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 5, 4, 3, 9, 8,
                    9, 1, 6, 2, 5};
      
        int n = arr.length;
      
        System.out.println(findRepeating(arr, n));
      
    }
}
// This code is contributed by 
// Smitha Dinesh Semwal.

Python3

# Python3 program to find the only 
# repeating element in an array 
# where elements are from 1 to n-1.
  
# Function to find repeted element
def findRepeating(arr, n):
  
    missingElement = 0
  
    # indexing based
    for i in range(0, n):
  
        element = arr[abs(arr[i])]
  
        if(element < 0):
            missingElement = arr[i]
            break
          
        arr[abs(arr[i])] = -arr[abs(arr[i])]
      
    return abs(missingElement)
  
# Driver code
arr = [5, 4, 3, 9, 8, 9, 1, 6, 2, 5]
n = len(arr)
print(findRepeating(arr, n))
  
# This code is contributed by Smitha Dinesh Semwal.

C＃

using System;
  
// C# program to find the only 
// repeating element in an array 
// where elements are from 1 to n-1. 
  
  
public class GFG
{
    // Function to find repeted element 
    public static int findRepeating(int[] arr, int n)
    {
        int missingElement = 0;
  
        // indexing based 
        for (int i = 0; i < n; i++)
        {
  
            int element = arr[Math.Abs(arr[i])];
  
            if (element < 0)
            {
                missingElement = arr[i];
                break;
            }
  
        arr[Math.Abs(arr[i])] = -arr[Math.Abs(arr[i])];
        }
  
    return Math.Abs(missingElement);
  
    }
  
    // Driver code 
    public static void Main(string[] args)
    {
        int[] arr = new int[] {5, 4, 3, 9, 8, 9,
                                1, 6, 2, 5};
  
        int n = arr.Length;
  
        Console.WriteLine(findRepeating(arr, n));
  
    }
}
  
// This code is contributed by Shrikant13

输出 ：

9

时间复杂度： O(n)
辅助空间： O(1)