On Mon, Jan 22, 2007 at 10:30:29AM +0000, Jon Dowland wrote:
> I think it's our duty to provide the most cunning/evil
> solution possible then :)
Probably not all that evil or cunning by most people's
standards, but here's my solution. I tried to do a
continuation-passing-style tail recursive thing, but my C
skills aren't up to it.
--
Jon Dowland
#include <stdio.h>
#include <stdlib.h>
/*
* find the number of unique values in an array
*/
typedef enum { false, true, } boolean;
struct node {
struct node * next;
int val;
};
boolean in_list(const int val, const struct node *list) {
if(NULL == list) return false;
return val == list->val || in_list(val, list->next);
}
struct node * add_to_list(const int val, struct node *list) {
if(NULL == list) {
list = malloc(sizeof(struct node));
if(!list) {
fprintf(stderr,"error allocating list node\n");
exit(1);
}
list->val = val;
list->next = NULL;
} else list->next = add_to_list(val, list->next);
return list;
}
int list_size(const struct node *list) {
if(NULL == list) return 0;
return 1 + list_size(list->next);
}
int main(int argc, char **argv) {
int i;
struct node *p, *list = NULL;
if(argc < 2) {
fprintf(stderr, "usage: %s number [ number ... ]\n", argv[0]);
exit(1);
}
/* populate the unique list */
for(i = 1; i < argc; ++i) {
int val = atoi(argv[i]);
if(!in_list(val, list)) {
list = add_to_list(val, list);
}
}
/* loop and print out the list */
for(p = list; NULL != p; p = p->next) {
printf("%d ", p->val);
}
printf("\n");
/* now for the new array */
printf("number of unique numbers: %d\n", list_size(list));
return 0;
}
