检查是否存在(i，j)以使arr [i]！= arr [j]并且arr [arr [i]]等于arr [arr [j]]

给定数组A []。任务是确定是否可以选择两个索引“ i”和“ j”以使满足以下条件：

1. A[i] is not equal to A[j].
2. A[A[i]] is equal to A[A[j]].

注意：数组中元素的值小于N的值，即，对于每个i，arr [i] 例子：

Input: N = 4, A[] = {1, 1, 2, 3}
Output: Yes
As A[3] != to A[1] but A[A[3]] == A[A[1]]

Input: N = 4, A[] = {2, 1, 3, 3}
Output: No
As A[A[3]] == A[A[4]] but A[3] == A[4]

方法：

1. 通过运行两个循环开始遍历数组Arr []。
2. 变量i指向索引0，变量j指向i的下一个。
3. 如果Arr [i]不等于Arr [j]，则检查Arr [Arr [i] – 1]是否等于Arr [Arr [j] – 1]。如果是，则返回true。
   否则，还检查Arr [Arr [i] -1]和Arr [Arr [j] – 1]是否有其他索引。
4. 重复上述步骤，直到遍历所有元素/索引。
5. 如果找不到这样的索引，则返回false。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
 
// Function that will tell whether
// such Indices present or Not.
bool checkIndices(int Arr[], int N)
{
 
    for (int i = 0; i < N - 1; i++) {
        for (int j = i + 1; j < N; j++) {
 
            // Checking 1st condition i.e whether
            // Arr[i] equal to Arr[j] or not
            if (Arr[i] != Arr[j]) {
 
                // Checking 2nd condition i.e whether
                // Arr[Arr[i]] equal to Arr[Arr[j]] or not.
                if (Arr[Arr[i] - 1] == Arr[Arr[j] - 1])
                    return true;
            }
        }
    }
 
    return false;
}
 
// Driver Code
int main()
{
    int Arr[] = { 3, 2, 1, 1, 4 };
    int N = sizeof(Arr) / sizeof(Arr[0]);
 
    // Calling function.
    checkIndices(Arr, N) ? cout << "Yes"
                         : cout << "No";
 
    return 0;
}

Java

// Java implementation of the above approach
 
// Function that calculates marks.
class GFG
{
    static boolean checkIndices(int Arr[], int N)
    {
   
        for (int i = 0; i < N - 1; i++) {
            for (int j = i + 1; j < N; j++) {
   
                // Checking 1st condition i.e whether
                // Arr[i] equal to Arr[j] or not
                if (Arr[i] != Arr[j]) {
     
                    // Checking 2nd condition i.e whether
                    // Arr[Arr[i]] equal to Arr[Arr[j]] or not.
                    if (Arr[Arr[i] - 1] == Arr[Arr[j] - 1])
                        return true;
                }
            }
        }
        return false;
    }
 
    // Driver code
    public static void main(String args[])
    {
        int Arr[] = { 3, 2, 1, 1, 4 };
        int N = Arr.length;
         
        if(checkIndices(Arr, N))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
 
// This code is Contributed by
// Naman_Garg

Python 3

# Python 3 implementation of the
# above approach
 
# Function that will tell whether
# such Indices present or Not.
def checkIndices(Arr, N):
 
    for i in range(N - 1):
        for j in range(i + 1, N):
 
            # Checking 1st condition i.e whether
            # Arr[i] equal to Arr[j] or not
            if (Arr[i] != Arr[j]):
 
                # Checking 2nd condition i.e whether
                # Arr[Arr[i]] equal to Arr[Arr[j]] or not.
                if (Arr[Arr[i] - 1] == Arr[Arr[j] - 1]):
                    return True
 
    return False
 
# Driver Code
if __name__ == "__main__":
     
    Arr = [ 3, 2, 1, 1, 4 ]
    N =len(Arr)
 
    # Calling function.
    if checkIndices(Arr, N):
        print("Yes") 
    else:
        print("No")
 
# This code is contributed by ita_c

C#

// C# implementation of the above approach
using System;
 
class GFG
{
     
// Function that calculates marks.
static bool checkIndices(int []Arr, int N)
{
    for (int i = 0; i < N - 1; i++)
    {
        for (int j = i + 1; j < N; j++)
        {
 
            // Checking 1st condition i.e whether
            // Arr[i] equal to Arr[j] or not
            if (Arr[i] != Arr[j])
            {
 
                // Checking 2nd condition i.e
                // whether Arr[Arr[i]] equal
                // to Arr[Arr[j]] or not.
                if (Arr[Arr[i] - 1] == Arr[Arr[j] - 1])
                    return true;
            }
        }
    }
    return false;
}
 
// Driver code
static public void Main ()
{
    int []Arr = { 3, 2, 1, 1, 4 };
    int N = Arr.Length;
     
    if(checkIndices(Arr, N))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is Contributed by Sachin

Javascript

 
// Javascript implementation of the above approach
 
// Function that will tell whether
// such Indices present or Not.
function checkIndices(Arr, N)
{
 
    for (var i = 0; i < N - 1; i++) {
        for (var j = i + 1; j < N; j++) {
 
            // Checking 1st condition i.e whether
            // Arr[i] equal to Arr[j] or not
            if (Arr[i] != Arr[j]) {
 
                // Checking 2nd condition i.e whether
                // Arr[Arr[i]] equal to Arr[Arr[j]] or not.
                if (Arr[Arr[i] - 1] == Arr[Arr[j] - 1])
                    return true;
            }
        }
    }
 
    return false;
}
 
// Driver Code
var Arr = [ 3, 2, 1, 1, 4 ];
var N = Arr.length;
 
// Calling function.
checkIndices(Arr, N) ? document.write( "Yes")
                    : document.write( "No");
 
 

输出：

Yes

时间复杂度： O(N ² )

辅助空间： O(1)