Write out a function that prints out all the permutations of
a string.
For example, abc would give you abc, acb, bac, bca, cab,
cba. You can assume that all the characters will be unique.
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Answer Posted / raghuram.a
/*guys..I have implemented Jhonson trotter algorithm..u can
print permutations of 123..n. implement the same for strings!*/
#include<iostream.h>
#include<conio.h>
int min(int a[10],int n)
{
int i,m=1;
for(i=2;i<=n;i++)
{
if(a[m]>a[i])
m=i;
}
return m;
}
void swap(int &a,int &b)
{
int t;
t=a;
a=b;
b=t;
}
int main()
{
int i,j,k,n,flag=0,l,m;
int d[100],a[100];
clrscr();
cout<<"\n\nenter n:\n\n";
cin>>n;
for(i=1;i<=n;i++)
a[i]=i;
for(i=1;i<=n;i++)
d[i]=i-1;
cout<<"\n\npermutations generated for integer 12...n
are:\n\n";
for(i=1;i<=n;i++) //display the given no.
cout<<a[i];
cout<<"\t";
do
{
flag=0;
k=min(a,n);
for(i=1;i<=n;i++)
{
if((a[i]>a[k])&&(d[i]!=0)&&(a[d[i]]<a[i])) //find
mobile integer.
k=i;
}
if(a[k]==1)
break;
l=d[k]; //copy of direction of mobile integer.
m=a[k]; //copy of mobile integer.
if(d[k]==k+1&&d[k+1]==k) //swap directions.
{
if(d[k]==n)
{ d[k+1]=0;
d[k]=k-1;
}
else if(d[k+1]==1)
{
d[k]=0;
d[k+1]=k+2;
}
else
{ d[k]=k-1;
d[k+1]=k+2;
}
}
else if(d[k]==k-1&&d[k-1]==k)
{
d[k]=k+1;
d[k-1]=k-2;
}
/*cout<<d[1]<<d[2]<<d[3]<<d[4];
cout<<"\t";
if u want to know directions of integers.
*/
swap(a[k],a[l]); //swap mobile integer and integer it is
pointing to.
for(i=1;i<=n;i++)
cout<<a[i]; //display the number.
cout<<"\t";
for(i=1;i<=n;i++) //reverse directions of integers
greater than
//mobile integer.
{
if(a[i]>m)
{
if(d[i]==0&&i==n)
d[i]=i-1;
else if(d[i]<i)
d[i]=i+1;
else if(d[i]>i)
d[i]=i-1;
}
}
for(i=1;i<n;i++) //check whether a mobile integer exist or not.
{
if(a[i]<a[i+1])
{
if(d[i+1]!=0)
flag=1;
}
else if(a[i]>a[i+1])
{
if(d[i]!=0)
flag=1;
}
}
}
while(flag==1); //if no mobile
integer(flag=0)terminate the program.
getch();
return 0;
}
| Is This Answer Correct ? | 4 Yes | 3 No |
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