在 O(1) 空间中查找具有元素 0 到 N-1 的常量数组中的重复项

给定一个由 n 个元素组成的常量数组，其中包含从 1 到 n-1 的元素，其中任何一个数字出现任意次数。在 O(n) 中找到这些重复数字中的任何一个，并且只使用恒定的内存空间。
例子：

Input : arr[] = {1, 2, 3, 4, 5, 6, 3}
Output : 3

由于给定的数组是下面文章中给出的常量方法将不起作用。

1. 在 O(n) 时间和 O(1) 额外空间内查找重复项 |设置 1
2. 在 O(n) 中的数组中重复并使用 O(1) 额外空间 |第 2 组

1. 我们从 0 开始取两个变量 i & j
2. 我们将循环运行，直到我到达最后一个元素或找到重复的元素
3. 我们将预先增加 j 值，以便我们可以将 elem 与下一个 elem 进行比较
4. 如果我们没有找到 elem，我们将增加 i，因为 j 将指向最后一个 elem，然后用 i 重新定位 j
   

Java

// Java program to find a duplicate
// element in an array with values in
// range from 0 to n-1.
import java.io.*;
import java.util.*;
 
public class GFG {
     
    // function to find one duplicate
    static int findduplicate(int []a, int n)
    {
         
        int i=0,j=0;
        while(i

---

Java

// Java program to find a duplicate
// element in an array with values in
// range from 0 to n-1.
import java.io.*;
import java.util.*;
 
public class GFG {
      
    // function to find one duplicate
    static int findduplicate(int []arr, int n)
    {
          
        // return -1 because in these cases
        // there can not be any repeated element
        if (n <= 1)
            return -1;
      
        // initialize fast and slow
        int slow = arr[0];
        int fast = arr[arr[0]];
      
        // loop to enter in the cycle
        while (fast != slow)
        {
      
            // move one step for slow
            slow = arr[slow];
      
            // move two step for fast
            fast = arr[arr[fast]];
        }
      
        // loop to find entry
        // point of the cycle
        fast = 0;
        while (slow != fast)
        {
            slow = arr[slow];
            fast = arr[fast];
        }
        return slow;
    }
      
    // Driver Code
    public static void main(String args[])
    {
        int []arr = {1, 2, 3, 4, 5, 6, 3};
        int n = arr.length;
        System.out.print(findduplicate(arr, n));
    }
}
  
// This code is contributed by
// Manish Shaw (manishshaw1)

Python 3

# Python 3 program to find a duplicate
# element in an array with values in
# range from 0 to n-1.
 
# function to find one duplicate
def findduplicate(arr, n):
 
    # return -1 because in these cases
    # there can not be any repeated element
    if (n <= 1):
        return -1
 
    # initialize fast and slow
    slow = arr[0]
    fast = arr[arr[0]]
 
    # loop to enter in the cycle
    while (fast != slow) :
 
        # move one step for slow
        slow = arr[slow]
 
        # move two step for fast
        fast = arr[arr[fast]]
 
    # loop to find entry point of the cycle
    fast = 0
    while (slow != fast):
        slow = arr[slow]
        fast = arr[fast]
    return slow
 
# Driver Code
if __name__ == "__main__":
     
    arr = [1, 2, 3, 4, 5, 6, 3 ]
    n = len(arr)
    print(findduplicate(arr, n))
 
# This code is contributed by ita_c

C#

// C# program to find a duplicate
// element in an array with values in
// range from 0 to n-1.
using System;
using System.Collections.Generic;
class GFG {
     
    // function to find one duplicate
    static int findduplicate(int []arr, int n)
    {
         
        // return -1 because in these cases
        // there can not be any repeated element
        if (n <= 1)
            return -1;
     
        // initialize fast and slow
        int slow = arr[0];
        int fast = arr[arr[0]];
     
        // loop to enter in the cycle
        while (fast != slow)
        {
     
            // move one step for slow
            slow = arr[slow];
     
            // move two step for fast
            fast = arr[arr[fast]];
        }
     
        // loop to find entry
        // point of the cycle
        fast = 0;
        while (slow != fast)
        {
            slow = arr[slow];
            fast = arr[fast];
        }
        return slow;
    }
     
    // Driver Code
    public static void Main()
    {
        int []arr = {1, 2, 3, 4, 5, 6, 3};
        int n = arr.Length;
        Console.Write(findduplicate(arr, n));
    }
}
 
// This code is contributed by
// Manish Shaw (manishshaw1)

Javascript

 
    // Javascript program to find a duplicate
    // element in an array with values in
    // range from 0 to n-1.
     
    // function to find one duplicate
    function findduplicate(arr, n)
    {
           
        // return -1 because in these cases
        // there can not be any repeated element
        if (n <= 1)
            return -1;
       
        // initialize fast and slow
        let slow = arr[0];
        let fast = arr[arr[0]];
       
        // loop to enter in the cycle
        while (fast != slow)
        {
       
            // move one step for slow
            slow = arr[slow];
       
            // move two step for fast
            fast = arr[arr[fast]];
        }
       
        // loop to find entry
        // point of the cycle
        fast = 0;
        while (slow != fast)
        {
            slow = arr[slow];
            fast = arr[fast];
        }
        return slow;
    }
     
    let arr = [1, 2, 3, 4, 5, 6, 3];
    let n = arr.length;
    document.write(findduplicate(arr, n));
     

输出：

3

时间复杂度： O(n)
辅助空间： O(1)