在整数数组中找到第一个重复元素

给定一个整数数组，在其中找到第一个重复元素。我们需要找到多次出现并且第一次出现的索引最小的元素。

例子：

Input:  arr[] = {10, 5, 3, 4, 3, 5, 6}
Output: 5 [5 is the first element that repeats]

Input:  arr[] = {6, 10, 5, 4, 9, 120, 4, 6, 10}
Output: 6 [6 is the first element that repeats]

一个简单的解决方案是使用两个嵌套循环。外循环一个接一个地选择一个元素，内循环检查该元素是否重复。找到重复的元素后，我们将中断循环并打印该元素。该解决方案的时间复杂度为O(n ² )

我们可以使用排序来解决O(nLogn)时间中的问题。以下是详细步骤。
1)将给定的数组复制到辅助数组temp []。
2)使用O(nLogn)时间排序算法对临时数组进行排序。
3)从左到右扫描输入阵列。对于每个元素，使用二进制搜索在temp []中计算其出现次数。一旦发现一个元素出现多次，我们将返回该元素。此步骤可以在O(nLogn)时间完成。
我们可以使用散列来平均解决O(n)时间。这个想法是从右到左遍历给定的数组，并在我们找到在右侧访问过的元素时更新最小索引。感谢Mohammad Shahid提出了此解决方案。

以下是该想法的实现。

C++

/* C++ program to find first repeating element in arr[] */
#include
using namespace std;
 
// This function prints the first repeating element in arr[]
void printFirstRepeating(int arr[], int n)
{
    // Initialize index of first repeating element
    int min = -1;
 
    // Creates an empty hashset
    set myset;
 
    // Traverse the input array from right to left
    for (int i = n - 1; i >= 0; i--)
    {
        // If element is already in hash set, update min
        if (myset.find(arr[i]) != myset.end())
            min = i;
 
        else   // Else add element to hash set
            myset.insert(arr[i]);
    }
 
    // Print the result
    if (min != -1)
        cout << "The first repeating element is " << arr[min];
    else
        cout << "There are no repeating elements";
}
 
// Driver method to test above method
int main()
{
    int arr[] = {10, 5, 3, 4, 3, 5, 6};
 
    int n = sizeof(arr) / sizeof(arr[0]);
    printFirstRepeating(arr, n);
}
//This article is contributed by Chhavi

Java

/* Java program to find first repeating element in arr[] */
import java.util.*;
 
class Main
{
    // This function prints the first repeating element in arr[]
    static void printFirstRepeating(int arr[])
    {
        // Initialize index of first repeating element
        int min = -1;
 
        // Creates an empty hashset
        HashSet set = new HashSet<>();
 
        // Traverse the input array from right to left
        for (int i=arr.length-1; i>=0; i--)
        {
            // If element is already in hash set, update min
            if (set.contains(arr[i]))
                min = i;
 
            else   // Else add element to hash set
                set.add(arr[i]);
        }
 
        // Print the result
        if (min != -1)
          System.out.println("The first repeating element is " + arr[min]);
        else
          System.out.println("There are no repeating elements");
    }
 
    // Driver method to test above method
    public static void main (String[] args) throws java.lang.Exception
    {
        int arr[] = {10, 5, 3, 4, 3, 5, 6};
        printFirstRepeating(arr);
    }
}

Python3

# Python3 program to find first repeating
# element in arr[]
 
# This function prints the first repeating
# element in arr[]
def printFirstRepeating(arr, n):
 
    # Initialize index of first repeating element
    Min = -1
 
    # Creates an empty hashset
    myset = dict()
 
    # Traverse the input array from right to left
    for i in range(n - 1, -1, -1):
     
        # If element is already in hash set,
        # update Min
        if arr[i] in myset.keys():
            Min = i
 
        else: # Else add element to hash set
            myset[arr[i]] = 1
     
    # Print the result
    if (Min != -1):
        print("The first repeating element is",
                                      arr[Min])
    else:
        print("There are no repeating elements")
 
# Driver Code
arr = [10, 5, 3, 4, 3, 5, 6]
 
n = len(arr)
printFirstRepeating(arr, n)
 
# This code is contributed by Mohit kumar 29

C#

using System;
using System.Collections.Generic;
 
/* C# program to find first repeating element in arr[] */
 
public class GFG
{
    // This function prints the first repeating element in arr[]
    public static void printFirstRepeating(int[] arr)
    {
        // Initialize index of first repeating element
        int min = -1;
 
        // Creates an empty hashset
        HashSet set = new HashSet();
 
        // Traverse the input array from right to left
        for (int i = arr.Length - 1; i >= 0; i--)
        {
            // If element is already in hash set, update min
            if (set.Contains(arr[i]))
            {
                min = i;
            }
 
            else // Else add element to hash set
            {
                set.Add(arr[i]);
            }
        }
 
        // Print the result
        if (min != -1)
        {
          Console.WriteLine("The first repeating element is " + arr[min]);
        }
        else
        {
          Console.WriteLine("There are no repeating elements");
        }
    }
 
    // Driver method to test above method
 
    public static void Main(string[] args)
    {
        int[] arr = new int[] {10, 5, 3, 4, 3, 5, 6};
        printFirstRepeating(arr);
    }
}
 
// This code is contributed by Shrikant13

输出

The first repeating element is 5

另一种方法：
如果要执行此操作而不使用任何其他数据结构。仅使用数组也可以解决该问题。请参阅下面的方法。

C++

/* C++ program to find first
repeating element in arr[] */
#include 
using namespace std;
 
// This function prints the
// first repeating element in arr[]
void printFirstRepeating(int arr[], int n)
{
     
    // This will set k=1, if any
    // repeating element found
    int k = 0;
 
    // max = maximum from (all elements & n)
    int max = n;
    for (int i = 0; i < n; i++)
        if (max < arr[i])
            max = arr[i];
 
    // Array a is for storing
    // 1st time occurence of element
    // initialized by 0
    int a[max + 1] = {};
 
    // Store 1 in array b
    // if element is duplicate
    // initialized by 0
    int b[max + 1] = {};
 
    for (int i = 0; i < n; i++)
    {
     
        // Duplicate element found
        if (a[arr[i]])
        {
            b[arr[i]] = 1;
            k = 1;
            continue;
        }
        else
            // storing 1st occurence of arr[i]
            a[arr[i]] = i;
    }
 
    if (k == 0)
        cout << "No repeating element found" << endl;
    else
    {
        int min = max + 1;
       
        // trace array a & find repeating element
        // with min index
        for (int i = 0; i < max + 1; i++)
            if (a[i] && min > a[i] && b[i])
                min = a[i];
        cout << arr[min];
    }
    cout << endl;
}
 
// Driver method to test above method
int main()
{
    int arr[] = { 10, 5, 3, 4, 3, 5, 6 };
 
    int n = sizeof(arr) / sizeof(arr[0]);
    printFirstRepeating(arr, n);
}

Java

/* Java program to find first
repeating element in arr[] */
public class GFG
{
 
  // This function prints the
  // first repeating element in arr[]
  static void printFirstRepeating(int[] arr, int n)
  {
 
    // This will set k=1, if any
    // repeating element found
    int k = 0;
 
    // max = maximum from (all elements & n)
    int max = n;
    for (int i = 0; i < n; i++)
      if (max < arr[i])
        max = arr[i];
 
    // Array a is for storing
    // 1st time occurence of element
    // initialized by 0
    int[] a = new int[max + 1];
 
    // Store 1 in array b
    // if element is duplicate
    // initialized by 0
    int[] b = new int[max + 1];
    for (int i = 0; i < n; i++)
    {
 
      // Duplicate element found
      if (a[arr[i]] != 0)
      {
        b[arr[i]] = 1;
        k = 1;
        continue;
      }
      else
        // storing 1st occurence of arr[i]
        a[arr[i]] = i;
    }
 
    if (k == 0)
      System.out.println("No repeating element found");
    else
    {
      int min = max + 1;
 
      // trace array a & find repeating element
      // with min index
      for (int i = 0; i < max + 1; i++)
        if (a[i] != 0 && min > a[i] && b[i] != 0)
          min = a[i];
      System.out.print(arr[min]);
    }
    System.out.println();
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int[] arr = { 10, 5, 3, 4, 3, 5, 6 };
 
    int n = arr.length;
    printFirstRepeating(arr, n);
  }
}
 
// This code is contributed by divyesh072019

Python3

# Python3 program to find first 
# repeating element in arr[]
 
# This function prints the 
# first repeating element in arr[]
def printFirstRepeating(arr, n):
     
    # This will set k=1, if any 
    # repeating element found
    k = 0
 
    # max = maximum from (all elements & n)
    max = n
     
    for i in range(n):
        if (max < arr[i]):
            max = arr[i]
     
    # Array a is for storing 
    # 1st time occurence of element
    # initialized by 0
    a = [0 for i in range(max + 1)]
 
    # Store 1 in array b 
    # if element is duplicate
    # initialized by 0
    b = [0 for i in range(max + 1)]
 
    for i in range(n):
 
        # Duplicate element found
        if (a[arr[i]]):
            b[arr[i]] = 1
            k = 1
            continue
        else:
 
            # Storing 1st occurence of arr[i]
            a[arr[i]] = i
 
    if (k == 0):
        print("No repeating element found")
    else:
        min = max + 1
 
        for i in range(max + 1):
             
            # Trace array a & find repeating
            # element with min index
            if (a[i] and (min > (a[i])) and b[i]):
                min = a[i]
                 
        print(arr[min])
 
# Driver code
arr = [ 10, 5, 3, 4, 3, 5, 6 ]
n = len(arr)
 
printFirstRepeating(arr, n)
 
# This code is contributed by avanitrachhadiya2155

C＃

/* C# program to find first
repeating element in arr[] */
using System;
class GFG
{
     
    // This function prints the
    // first repeating element in arr[]
    static void printFirstRepeating(int[] arr, int n)
    {
          
        // This will set k=1, if any
        // repeating element found
        int k = 0;
      
        // max = maximum from (all elements & n)
        int max = n;
        for (int i = 0; i < n; i++)
            if (max < arr[i])
                max = arr[i];
      
        // Array a is for storing
        // 1st time occurence of element
        // initialized by 0
        int[] a = new int[max + 1];
      
        // Store 1 in array b
        // if element is duplicate
        // initialized by 0
        int[] b = new int[max + 1];
      
        for (int i = 0; i < n; i++)
        {
          
            // Duplicate element found
            if (a[arr[i]] != 0)
            {
                b[arr[i]] = 1;
                k = 1;
                continue;
            }
            else
                // storing 1st occurence of arr[i]
                a[arr[i]] = i;
        }
      
        if (k == 0)
            Console.WriteLine("No repeating element found");
        else
        {
            int min = max + 1;
            
            // trace array a & find repeating element
            // with min index
            for (int i = 0; i < max + 1; i++)
                if ((a[i] != 0) && min > a[i] && (b[i] != 0))
                    min = a[i];
            Console.Write(arr[min]);
        }
        Console.WriteLine();
    }
 
  // Driver code
  static void Main()
  {
    int[] arr = { 10, 5, 3, 4, 3, 5, 6 };
  
    int n = arr.Length;
    printFirstRepeating(arr, n);
  }
}
 
// This code is contributed by divyeshrabadiya07.

输出

5

时间复杂度： O(n)。

使用O(n)辅助空间和O(n)时间复杂度的另一种方法：

Java

/*package whatever //do not write package name here */
 
import java.io.*;
class GFG {
    public static int firstRepeated(int[] arr, int n)
    {
        int max = 0;
        for (int x = 0; x < n; x++) {
            if (arr[x] > max) {
                max = arr[x];
            }
        }
        int temp[]
            = new int[max + 1]; // the idea is to use
                                // temporary array as hashmap
        Arrays.fill(temp, 0);
 
        for (int x = 0; x < n; x++) {
            int num = arr[x];
            temp[num]++;
        }
 
        for (int x = 0; x < n; x++) {
            int num = arr[x];
            if (temp[num] > 1) {
                return x;
            }
        }
 
        return -1; // if no repeating element found
    }
    public static void main(String[] args)
    {
        int[] arr = { 10, 5, 3, 4, 3, 5, 6 };
 
        int n = arr.length;
        int index = firstRepeated(
            arr,
            arr.length); // index Of 1st repeating number
        if (index != -1) {
            System.out.println("1st Repeatig Number is  "
                               + arr[index]);
        }
        else {
            System.out.println("No Repeating Number Found");
        }
    }
}

使用Python内置函数的另一种方法：

Python3

# Python3 program to find first
# repeating element in arr[]
 
# This function prints the
# first repeating element in arr[]
def printFirstRepeating(a, n):
    for i in range(len(a)):
        if a.count(a[i]) > 1:
            return a[i]
    return "there is no repetition number"
 
 
# Driver code
arr = [10, 5, 3, 4, 3, 5, 6]
n = len(arr)
print(printFirstRepeating(arr, n))
 
# This code is contributed by karthikeyakumarnallam

输出

5