找到给定序列的第K个位置元素

鉴于两个整数N和K，任务是在K找到元素位置^i，如果从1到N的所有奇数为了增加随后从1所有的偶数到N递增的顺序都写下来。
例子：

Input: N = 10, K = 3 
Output: 5 
The required sequence is 1, 3, 5, 7, 9, 2, 4, 6, 8 and 10.
Input: N = 7, K = 7 
Output: 6 
 

方法：已知第N^个偶数由2 * K给出，第N^个奇数由2 * K – 1给出。但是，由于此处的偶数写在(N +1)/ 2个奇数之后。因此，^第K^个偶数由2 *(K –(N + 1)/ 2)给出，奇数保持与2 * K – 1相同。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the kth number
// from the required sequence
int kthNum(int n, int k)
{
 
    // Count of odd integers
    // in the sequence
    int a = (n + 1) / 2;
 
    // kth number is even
    if (k > a)
        return (2 * (k - a));
 
    // It is odd
    return (2 * k - 1);
}
 
// Driver code
int main()
{
    int n = 7, k = 7;
 
    cout << kthNum(n, k);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
 
// Function to return the kth number
// from the required sequence
static int kthNum(int n, int k)
{
 
    // Count of odd integers
    // in the sequence
    int a = (n + 1) / 2;
 
    // kth number is even
    if (k > a)
        return (2 * (k - a));
 
    // It is odd
    return (2 * k - 1);
}
 
// Driver code
public static void main(String []args)
{
    int n = 7, k = 7;
 
    System.out.println(kthNum(n, k));
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach
 
# Function to return the kth number
# from the required sequence
def kthNum(n, k) :
 
    # Count of odd integers
    # in the sequence
    a = (n + 1) // 2;
 
    # kth number is even
    if (k > a) :
        return (2 * (k - a));
 
    # It is odd
    return (2 * k - 1);
 
# Driver code
if __name__ == "__main__" :
 
    n = 7; k = 7;
 
    print(kthNum(n, k));
 
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
     
class GFG
{
 
// Function to return the kth number
// from the required sequence
static int kthNum(int n, int k)
{
 
    // Count of odd integers
    // in the sequence
    int a = (n + 1) / 2;
 
    // kth number is even
    if (k > a)
        return (2 * (k - a));
 
    // It is odd
    return (2 * k - 1);
}
 
// Driver code
public static void Main(String []args)
{
    int n = 7, k = 7;
 
    Console.WriteLine(kthNum(n, k));
}
}
 
// This code is contributed by PrinciRaj1992

Javascript

 
// Javascript implementation of the approach
 
// Function to return the kth number
// from the required sequence
function kthNum(n, k)
{
    // Count of odd integers
    // in the sequence
    var a = (n + 1) / 2;
 
    // kth number is even
    if (k > a)
        return (2 * (k - a));
 
    // It is odd
    return (2 * k - 1);
}
 
// Driver code
var n = 7, k = 7;
document.write(kthNum(n, k));
 

输出：

6

时间复杂度： O(1)