i have posted similar problem here
http://stackoverflow.com/questions/2920315/permutation-of-array
but i want following we know that with length n there is n! possible permutation from which one such that all element are in order they are in sorted variant so i want break permutation when
array is in order and print result but something is wrong i think that problem is repeated of permutation
here is my code
import java.util.*;
public class permut{
public static Random r=new Random();
public static void display(int a[],int n){
for (int i=0;i<n;i++){
System.out.println(a[i]);
}
}
public static void Permut(int a[],int n){
int j=0;
int k=0;
while (j<fact(n)){
int s=r.nextInt(n);
for (int i=0;i<n;i++){
k=a[i];
a[i]=a[s];
a[s]=k;
}
j++;
if (sorted(a,n))
display(a,n);
break;
}
}
public static void main(String[]args){
int a[]=new int[]{3,4,1,2};
int n=a.length;
Permut(a,n);
}
public static int fact(int n){
if (n==0 || (n==1) )
return 1;
return n*fact(n-1);
}
public static boolean sorted(int a[],int n ){
boolean flag=false;
for (int i=0;i<n-1;i++){
if (a[i]<a[i+1]){
flag=true;
}
else{
flag=false;
}
}
return flag;
}
}
can anybody help me? result is nothing
