CareerCup

1. By using recursion , we can solve this problem.
2. You need two Arrays one is Permuted array P[] which stores permutation numbers.
second Array is Presence Arraya Presence[] which tells whether a particular number is already included in P[] or not .

3. Compiled and tested Code for this.

#include<stdio.h>

#define TRUE 1
#define FALSE 0

void PrintArray(int *P, int n)
{
	int i=0;
	while(i<n)
	{
		printf(" %d , ", P[i] );
		i++;
	}
	printf("\n");
}

/* 
int P[]: Permutation Array which stores numbers permutation 

int Presence[]:  it store either 0 or 1 for all numbers 1 - n . 
if Presence[i] == 1, i'th number already there in P[] array. 
So we will not include that number again.

int Count: tells so far how many numbers included in Permuted Array. 
*/

void Permutations(int *P, int *Presence, int n, int count )
{
	if( count >= n ||  n <= 0)
	{
		return;
	}
	else
	{
		int i = 1;

		while(i <= n)
		{
			if( Presence[i] != TRUE )
			{
				Presence[i] = TRUE;
				P[count] = i;
				if( count + 1 == n )
				{
					PrintArray(P,n); // Print all elements in P[]
				}
				else
				{
					Permutations(P, Presence, n, count + 1 );
				}
				Presence[i] = FALSE;
				
			}
			i++;
		}
	}
}

int main()
{
	int n = 5; // 1 to 10 numbers ermutations 
	int P[n] ; // Permutation Array, which store numbers permutation
	int Presence[n]; // it store either 0 or 1 for all numbers 1 - n 
	int i =0;

	/* Intializing arrays */
	while( i < n )
	{
		P[i] = Presence[i] = 0;
		i++;
	}
 
	Permutations(P, Presence, n, 0 );

	return 0;
}

Is there any other better way ?

U can do it with single array...Call Perm(a,0).a has the integers 1 to n

Perm(int[] a,int start){
  if(start==a.length){
    Print a
    return;
  }
  int temp;
  for(int i=start;i<a.length;i++){
    swap(a[start],a[i])
    Perm(a,start+1)
    swap(a[start],a[i])
  }
}