给定整数N。任务是找到前N个自然数的置换,以使任意两个连续数之间的绝对差> 1 。如果不可能进行这种排列,则打印-1 。
例子:
Input: N = 5
Output: 5 3 1 4 2
Input: N = 3
Output: -1
方法:可能有许多这样的排列方式,但是最常见和贪婪的方法之一是以降序(或递增)顺序排列所有奇数,然后以递减(或递增)顺序排列所有偶数。请注意,如果N = 3或N = 2,那么将不可能有这种排列方式;如果N = 1,则序列将由单个元素组成,即1 。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to print the required permutation
void arrange(int N)
{
if (N == 1) {
cout << "1";
return;
}
// No permutation is possible
// satisfying the given condition
if (N == 2 || N == 3) {
cout << "-1";
return;
}
// Maximum even and odd elements < N
int even = -1, odd = -1;
if (N % 2 == 0) {
even = N;
odd = N - 1;
}
else {
odd = N;
even = N - 1;
}
// Print all odd elements in decreasing order
while (odd >= 1) {
cout << odd << " ";
// Next element must be odd
odd = odd - 2;
}
// Print all even elements in decreasing order
while (even >= 2) {
cout << even << " ";
// Next element must be even
even = even - 2;
}
}
// Driver code
int main()
{
int N = 5;
arrange(N);
return 0;
}
Java
// Java implementation of the approach
class GFG
{
// Function to print the required
// permutation
static void arrange(int N)
{
if (N == 1)
{
System.out.println("1");
return;
}
// No permutation is possible
// satisfying the given condition
if (N == 2 || N == 3)
{
System.out.println("-1");
return;
}
// Maximum even and odd elements < N
int even = -1, odd = -1;
if (N % 2 == 0)
{
even = N;
odd = N - 1;
}
else
{
odd = N;
even = N - 1;
}
// Print all odd elements in
// decreasing order
while (odd >= 1)
{
System.out.print(odd);
System.out.print(" ");
// Next element must be odd
odd = odd - 2;
}
// Print all even elements in
// decreasing order
while (even >= 2)
{
System.out.print(even);
System.out.print(" ");
// Next element must be even
even = even - 2;
}
}
// Driver code
public static void main(String[] args)
{
int N = 5;
arrange(N);
}
}
// This code is contributed
// by Akanksha Rai
Python3
# Python3 implementation of the approach
# Function to print the required permutation
def arrange(N):
if (N == 1) :
print("1")
return
# No permutation is possible
# satisfying the given condition
if (N == 2 or N == 3) :
print("-1")
return
# Maximum even and odd elements < N
even = -1
odd = -1
if (N % 2 == 0):
even = N
odd = N - 1
else :
odd = N
even = N - 1
# Print all odd elements in
# decreasing order
while (odd >= 1):
print(odd, end = " ")
# Next element must be odd
odd = odd - 2
# Print all even elements in
# decreasing order
while (even >= 2):
print(even, end = " ")
# Next element must be even
even = even - 2
# Driver code
if __name__ == "__main__":
N = 5
arrange(N)
# This code is contributed by ita_c
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to print the required
// permutation
static void arrange(int N)
{
if (N == 1)
{
Console.WriteLine("1");
return;
}
// No permutation is possible
// satisfying the given condition
if (N == 2 || N == 3)
{
Console.WriteLine("-1");
return;
}
// Maximum even and odd elements < N
int even = -1, odd = -1;
if (N % 2 == 0)
{
even = N;
odd = N - 1;
}
else
{
odd = N;
even = N - 1;
}
// Print all odd elements in
// decreasing order
while (odd >= 1)
{
Console.Write(odd);
Console.Write(" ");
// Next element must be odd
odd = odd - 2;
}
// Print all even elements in
// decreasing order
while (even >= 2)
{
Console.Write(even);
Console.Write(" ");
// Next element must be even
even = even - 2;
}
}
// Driver code
public static void Main()
{
int N = 5;
arrange(N);
}
}
// This code is contributed
// by Shivi_Aggarwal
PHP
= 1)
{
echo $odd, " ";
// Next element must be odd
$odd = $odd - 2;
}
// Print all even elements in
// decreasing order
while ($even >= 2)
{
echo $even, " ";
// Next element must be even
$even = $even - 2;
}
}
// Driver code
$N = 5;
arrange($N);
// This code is contributed by Ryuga
?>
Javascript
输出:
5 3 1 4 2
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