遍历 1 到 N 的排列数组 K 步后查找索引

给定一个整数K和一个长度为N的索引数组arr[] ，其中包含范围 [1, N] 中的元素，任务是从索引 1 开始遍历数组K步后找到索引。
索引数组的遍历：在索引数组的遍历中，下一个要访问的索引是当前索引处的值。
例子：

Input: arr[] = {3, 2, 4, 1}, K = 5 
Output: 4 
Explanation: 
Traversal to the indices starting from index 1: 
1 => 3 => 4 => 1 => 3 => 4 
Finally, after K traversals the index is 4
Input: arr[] = {6, 5, 2, 5, 3, 2}, K = 2 
Output: 5 
 

方法：问题中的关键观察是，在遍历索引数组 N 次后，它会重复自己。因此，我们可以找到$K \% N$然后最后找到索引之后$K \% N$遍历。
下面是上述方法的实现：

C++

// C++ implementation to find the
// index after traversing the index
// array K times
 
#include
using namespace std;
 
// Function to find the index after
// traversing the index array K times
int findIndexAfterKTrav(vector arr,
                        int n, int k){
    k = k % n;
    int indi = 1;
     
    // Loop to traverse the index
    // array K times
    while (k){
        indi = arr[indi-1];
        k--;
    }
     
    return arr[indi-1];
}
 
// Driver Code
int main() {
    int n = 4, k = 5;
    vector arr{3, 2, 4, 1};
 
    // Function Call
    cout << findIndexAfterKTrav(arr, n, k);
    return 0;
}

Java

// Java implementation to find the
// index after traversing the index
// array K times
class GFG{
     
// Function to find the index after
// traversing the index array K times
public static int findIndexAfterKTrav(int[] arr,
                                      int n, int k)
{
    k = k % n;
    int indi = 1;
         
    // Loop to traverse the index
    // array K times
    while (k > 0)
    {
        indi = arr[indi - 1];
        k--;
    }
    return arr[indi - 1];
}
 
// Driver code
public static void main(String[] args)
{
    int n = 4, k = 5;
    int[] arr = { 3, 2, 4, 1 };
     
    // Function Call
    System.out.print(findIndexAfterKTrav(arr, n, k));
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 implementation to find the
# index after traversing the index
# array K times
 
# Function to find the index after
# traversing the index array K times
def findIndexAfterKTrav(arr, n, k):
     
    k = k % n;
    indi = 1;
 
    # Loop to traverse the index
    # array K times
    while (k > 0):
        indi = arr[indi - 1];
        k -= 1;
 
    return arr[indi - 1];
 
# Driver code
if __name__ == '__main__':
     
    n = 4;
    k = 5;
    arr = [ 3, 2, 4, 1 ];
 
    # Function Call
    print(findIndexAfterKTrav(arr, n, k));
 
# This code is contributed by Princi Singh

C#

// C# implementation to find the
// index after traversing the index
// array K times
using System;
 
class GFG{
      
// Function to find the index after
// traversing the index array K times
public static int findIndexAfterKTrav(int[] arr,
                                      int n, int k)
{
    k = k % n;
    int indi = 1;
          
    // Loop to traverse the index
    // array K times
    while (k > 0)
    {
        indi = arr[indi - 1];
        k--;
    }
    return arr[indi - 1];
}
  
// Driver code
public static void Main(String[] args)
{
    int n = 4, k = 5;
    int[] arr = { 3, 2, 4, 1 };
      
    // Function Call
    Console.Write(findIndexAfterKTrav(arr, n, k));
}
}
 
// This code is contributed by Princi Singh

Javascript

 
// Javascript implementation to find the
// index after traversing the index
// array K times
 
// Function to find the index after
// traversing the index array K times
function findIndexAfterKTrav(arr, n, k)
{
    k = k % n;
    var indi = 1;
     
    // Loop to traverse the index
    // array K times
    while (k){
        indi = arr[indi-1];
        k--;
    }
     
    return arr[indi-1];
}
 
// Driver Code
var n = 4, k = 5;
var arr = [3, 2, 4, 1];
 
// Function Call
document.write( findIndexAfterKTrav(arr, n, k));
 
// This code is contributed by itsok.

输出：

4

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