程序打印系列2，1，4，3，6，5，…。最多N个词

给定数字N，任务是打印以下模式，最多N个词：

2, 1, 4, 3, 6, 5, …. N terms 
 

例子：

Input: N = 4
Output: 2, 1, 4, 3

Explanation:
Nth Term = (N % 2 == 0) ? (N - 1) : (N + 1)
1st term = (1 % 2 == 0) ? (1 - 1) : (1 + 1)
         = (1 + 1)
         = 2
2nd term = (2 % 2 == 0) ? (2 - 1) : (2 + 1)
         = (2 - 1)
         = 1
3rd term = (3 % 2 == 0) ? (3 - 1) : (3 + 1)
         = (3 + 1)
         = 4
4th term = (4 % 2 == 0) ? (4 - 1) : (4 + 1)
         = (4 - 1)
         = 3
Therefore, Series = 2, 1, 4, 3

Input: N = 7
Output: 2, 1, 4, 3, 6, 5, 8

公式：

Nth Term = (N % 2 == 0) ? (N - 1) : (N + 1)

以下是上述问题的解决方案：

C++

// C++ program to print the series
// 2, 1, 4, 3, 6, 5, …. up to N terms
#include 
using namespace std;
 
// Function to print the series
void printPattern(int N)
{
 
    for (int i = 1; i <= N; i++) {
 
        // Find and print the ith term
        cout <<" "<<((i % 2 == 0) ? (i - 1) : (i + 1));
    }
}
 
// Driver code
int main()
{
 
    // Get the value of N
    int N = 10;
 
    // Print the Series
    printPattern(N);
 
    return 0;
}

Java

// Java Program to print the series
// 2, 1, 4, 3, 6, 5, …. up to N terms
 
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG
{
// Function to print the series
static void printPattern(int N)
{
   
    for (int i = 1; i <= N; i++) {
   
        // Find and print the ith term
        System.out.print(" "+((i % 2 == 0) ? (i - 1) : (i + 1)));
    }
}
   
// Driver code
public static void main(String args[])
{
   
    // Get the value of N
    int N = 10;
   
    // Print the Series
    printPattern(N);
   
}
}

Python3

# Python program to print the series
# 2, 1, 4, 3, 6, 5, …. up to N terms
 
# Function to print the series
def printPattern(N):
    for i in range(1, N + 1):
         
        # Find and print the ith term
        print(i - 1 if i % 2 == 0
                    else i + 1, end = " ")
 
# Driver code
N = 10
printPattern(N)
 
# This code is contributed by Shrikant13

C#

// C# program to print the series
// 2, 1, 4, 3, 6, 5, …. up to N terms
using System;
 
class GFG
{
// Function to print the series
public void printPattern(int N)
{
 
    for (int i = 1; i <= N; i++)
    {
 
        // Find and print the ith term
        int a = ((i % 2 == 0) ?
                      (i - 1) : (i + 1));
        Console.Write(" {0}", a);
    }
}
 
// Driver code
public static void Main()
{
    GFG g = new GFG();
     
    // Get the value of N
    int N = 10;
 
    // Print the Series
    g.printPattern(N);
}
}
 
// This code is contributed
// by Soumik

Javascript

// javascript program to print the series
// 2, 1, 4, 3, 6, 5, …. up to N terms
 
// Function to print the series
function printPattern( N)
{
 
    for (let i = 1; i <= N; i++)
    {
 
        // Find and print the ith term
        document.write(" " + ((i % 2 == 0) ? (i - 1) : (i + 1)));
    }
}
 
// Driver code
 
    // Get the value of N
    let N = 10;
 
    // Print the Series
    printPattern(N);
     
// This code is contributed by Rajput-Ji
 

输出：

2 1 4 3 6 5 8 7 10 9