@ravu
I have mentioned how next permutation works.. Also i have given an
example in another post. Please go thru the link..
On Dec 28, 10:47 pm, Gene <gene.ress...@gmail.com> wrote:
> Don't forget repeats. The string "aaa" has only one permutation.
>
> A related interesting question is how to write a permutation
> iterator. Here is the interface:
>
> typedef struct {
> // your code here
>
> } PERMUTATION_ITERATOR;
>
> /* Initialize the given permutation iterator with the string of n
> characters
> to be permuted. */
> void init_permutation_iterator(PERMUTATION_ITERATOR *iterator, char
> *str, int n)
> {
> // your code here
>
> }
>
> /* Get the next unique permutation of the initialization string into
> the
> given buffer. The first string returned is always the string
> provided
> to the initializer. Return true unless the permutation being
> returned
> is the last one. I.e. next time this function is called, it will
> "wrap" back
> to the initialization string. */
> int get_next_permutation(PERMUTATION_ITERATOR *iterator, char *buf)
> {
> // your code here
>
> }
>
> Use like this:
>
> {
> PERMUTATION_ITERATOR iterator[1];
> char s = "Cool or what?";
> char buf[100];
>
> init_permutation_iterator(iterator, s, strlen(s));
> while (get_next_permutation(iterator, buf))
> printf("%.*s\n", buf);
>
> }
>
> On Dec 28, 8:15 am, "Karthikeyan V.B" <kartmu...@gmail.com> wrote:
>
>
>
>
>
>
>
> > Hi,
>
> > Using Backtracking,
>
> > void swap(char* x,char* y)
> > {
> > char temp;
> > temp=*x;
> > *x=*y;
> > *y=temp;
>
> > }
>
> > void permute(char* a,int i,int n)
> > {
> > int j;
> > if(i==n)
> > printf("%s\n",a);
> > else
> > {
> > for(j=i;j<=n;j++)
> > {
> > swap((a+i),(a+j));
> > permute(a,i+1,n);
> > swap((a+i),(a+j));
>
> > }
> > }
> > }
>
> > But this takes O(n*n!) time
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