找到给定序列的第 k 个位置元素
原文:https://www . geeksforgeeks . org/find-the-kth-position-element-of-the-给定序列/
给定两个整数 N 和 K ，如果从 1 到 N 的所有奇数按递增顺序写下，然后从 1 到 N 的所有偶数按递增顺序写下，那么任务是在 K 第个位置找到元素。 举例:
输入: N = 10，K = 3 输出: 5 所需顺序为 1、3、5、7、9、2、4、6、8、10。 输入: N = 7，K = 7 输出: 6
进场:已知 N 第T5】偶数由 2 * K 给出， N 第T11】奇数由2 * K–1给出。但是由于这里偶数写在 (N + 1) / 2 奇数之后。因此， K th 偶数由2 (K–(N+1)/2)给出，奇数将保持与2 * K–1** 相同，以下是上述方法的实现:*
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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
java 描述语言
<script>
// 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));
</script>
Output:
6
时间复杂度: O(1)
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