#include "config.h"
#include <stdio.h>
/*
* Demonstration code for sorting a linked list.
*
* The algorithm used is Mergesort, because that works really well
* on linked lists, without requiring the O(N) extra space it needs
* when you do it on arrays.
*
* This code can handle singly and doubly linked lists, and
* circular and linear lists too. For any serious application,
* you'll probably want to remove the conditionals on `is_circular'
* and `is_double' to adapt the code to your own purpose.
*
*/
/*
* This file is copyright 2001 Simon Tatham.
*
* Permission is hereby granted, free of charge, to any person
* obtaining a copy of this software and associated documentation
* files (the "Software"), to deal in the Software without
* restriction, including without limitation the rights to use,
* copy, modify, merge, publish, distribute, sublicense, and/or
* sell copies of the Software, and to permit persons to whom the
* Software is furnished to do so, subject to the following
* conditions:
*
* The above copyright notice and this permission notice shall be
* included in all copies or substantial portions of the Software.
*
* THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND,
* EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES
* OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND
* NONINFRINGEMENT. IN NO EVENT SHALL SIMON TATHAM BE LIABLE FOR
* ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF
* CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN
* CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
* SOFTWARE.
*/
#include "listsort.h"
typedef unsigned char byte;
#include "word.h"
#ifdef TEST
static int cmp(const element *a, const element *b);
static int cmp(const element *a, const element *b) {
return a->i - b->i;
}
#endif
/*
* This is the actual sort function. Notice that it returns the new
* head of the list. (It has to, because the head will not
* generally be the same element after the sort.) So unlike sorting
* an array, where you can do
*
* sort(myarray);
*
* you now have to do
*
* list = listsort(mylist);
*/
element *listsort(element *list, fcn_compare *compare) {
element *p, *q, *e, *tail;
int insize, nmerges, psize, qsize, i;
/*
* Silly special case: if `list' was passed in as NULL, return
* NULL immediately.
*/
if (!list)
return NULL;
insize = 1;
while (1) {
p = list;
list = NULL;
tail = NULL;
nmerges = 0; /* count number of merges we do in this pass */
while (p) {
nmerges++; /* there exists a merge to be done */
/* step `insize' places along from p */
q = p;
psize = 0;
for (i = 0; i < insize; i++) {
psize++;
q = q->next;
if (!q) break;
}
/* if q hasn't fallen off end, we have two lists to merge */
qsize = insize;
/* now we have two lists; merge them */
while (psize > 0 || (qsize > 0 && q)) {
/* decide whether next element of merge comes from p or q */
if (psize == 0) {
/* p is empty; e must come from q. */
e = q; q = q->next; qsize--;
} else if (qsize == 0 || !q) {
/* q is empty; e must come from p. */
e = p; p = p->next; psize--;
} else if (compare(p,q) <= 0) {
/* First element of p is lower (or same);
* e must come from p. */
e = p; p = p->next; psize--;
} else {
/* First element of q is lower; e must come from q. */
e = q; q = q->next; qsize--;
}
/* add the next element to the merged list */
if (tail) {
tail->next = e;
} else {
list = e;
}
tail = e;
}
/* now p has stepped `insize' places along, and q has too */
p = q;
}
tail->next = NULL;
/* If we have done only one merge, we're finished. */
if (nmerges <= 1) /* allow for nmerges==0, the empty list case */
return list;
/* Otherwise repeat, merging lists twice the size */
insize *= 2;
}
}
/*
* Small test rig with three test orders. The list length 13 is
* chosen because that means some passes will have an extra list at
* the end and some will not.
*/
#ifdef TEST
int main(void) {
#define n 13
element k[n], *head, *p;
int order[][n] = {
{ 0,1,2,3,4,5,6,7,8,9,10,11,12 },
{ 6,2,8,4,11,1,12,7,3,9,5,0,10 },
{ 12,11,10,9,8,7,6,5,4,3,2,1,0 },
};
unsigned int i, j;
for (j = 0; j < n; j++)
k[j].i = j;
listsort(NULL, cmp, 0, 0);
for (i = 0; i < sizeof(order)/sizeof(*order); i++) {
int *ord = order[i];
head = &k[ord[0]];
for (j = 0; j < n; j++) {
if (j == n-1)
k[ord[j]].next = NULL;
else
k[ord[j]].next = &k[ord[j+1]];
}
printf("before:");
p = head;
do {
printf(" %d", p->i);
p = p->next;
} while (p != NULL);
printf("\t");
head = listsort(head, cmp);
printf(" after:");
p = head;
do {
printf(" %d", p->i);
p = p->next;
} while (p != NULL);
printf("\n");
}
printf("\n");
return 0;
}
#endif