找出 1 到 n-1 之间唯一的重复元素

我们得到一个大小为 n 的数组 arr[]。数字是从 1 到 (n-1) 的随机顺序。该数组只有一个重复元素。我们需要找到重复的元素。

例子：

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

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

方法一（简单）我们使用两个嵌套循环。外循环遍历所有元素，内循环检查外循环选取的元素是否出现在其他任何地方。

时间复杂度： O(n*n)

方法二（使用求和公式）：我们知道前n-1个自然数的和是(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缺失的元素。

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)
导致大型数组溢出。

方法三（Use Hashing）：使用哈希表来存储访问过的元素。如果看到的元素再次出现，我们就返回它。

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

蟒蛇3

# 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）：这个想法是基于x ^ x = 0和x ^ y = y ^ x的事实。
1) 计算从 1 到 n-1 的元素的异或。
2) 计算数组元素的异或。
3）以上两个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.

蟒蛇3

# 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.

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.对于每一个索引访问a[index]，如果它是正的改变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 repeated 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 repeated 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.

蟒蛇3

# Python3 program to find the only
# repeating element in an array
# where elements are from 1 to n-1.
 
# Function to find repeated 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.

Javascript

 
       // JavaScript program for the above approach;
 
       // Function to find repeated element
       function findRepeating(arr, n) {
           let missingElement = 0;
 
           // indexing based
           for (let i = 0; i < n; i++) {
 
               let 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
 
       let arr = [5, 4, 3, 9, 8,
           9, 1, 6, 2, 5];
 
       let n = arr.length;
 
       document.write(findRepeating(arr, n));
  // This code is contributed by Potta Lokesh
   

输出 ：

9

时间复杂度： O(n)
辅助空间： O(1)

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