Komplette Elemente einer Liste vertauschen

[Siderius]

Hallo zusammen,

ich hänge hier gerade an einer Sortierung einer Liste... Das funktioniert auch soweit, nur ist unser Dozent damit nicht einverstanden, da wir die Inhalte der Elemente und nicht die Elemente an sich vertauscht haben...

Nun die Frage:

Wie kann man das realisieren.

Anbei unser bisheriger Code:

struct ListenElement * SortList (struct ListenElement * liste)
{
	struct ListenElement *pointer1, *pointer2;
	struct ListenElement Temp;

	if (liste!=NULL)
	{
		pointer1 = liste;

		while(pointer1 != NULL)
		{
			pointer2 = pointer1->next;
			while(pointer2 != NULL)
			{
				if (strcmp(pointer1->Name,pointer2->Name) > 0)
				{
					/*Temp.next = pointer2;			
					pointer2 = pointer1;
					pointer1 = Temp.next;*/

					strcpy(Temp.Name, pointer2->Name);
					Temp.Alter = pointer2->Alter;

					strcpy(pointer2->Name, pointer1->Name);
					pointer2->Alter = pointer1->Alter;

					strcpy(pointer1->Name, Temp.Name);
					pointer1->Alter = Temp.Alter;

				}

				pointer2 = pointer2->next;
			}
			pointer1 = pointer1->next;
		}

		return liste;
	}
	else
	{
	printf("Keine Elemente zum sortieren vorhanden!");
	return;
	}
}

[Siderius]

Wollte nur den Beitrag eben übernehmen, nicht dass man meint ich würde mich einfach so hier rumtreiben, sondern will effektiv mich am Forum beteiligen...

[c_newbie]

Wie sieht dein struct ListenElement aus? Ist das ne einfache oder doppelt verkettete Liste?

[Siderius]

Die einzelnen Elemente sind Inhalt einer einfach verketteten Liste...

Die Elemente sehen folgendermaßne aus:

struct ListenElement
{
	char Name [20];
	int Alter;
	struct ListenElement *next;
};

[c_newbie]

direkt sortiert einfügen dürft ihr nicht? wenn du dann noch einen prev pointer
anlegst brauchst nie wieder sortieren

[Siderius]

Nein leider dürfen wir das nicht...

Wir sollen
a) Liste unsortiert erstelle
a1) Liste darstellen
b) einen eintrag ausfindig machen
c) diesen eintrag löschen
(bis hier bin ich und es klappt alles)
d) Liste durch vertauschen der Elemente sortieren
(dafür die Frage)
e) Komplette Liste löschen...

Auf die Idee es von Anfang an zu sortieren hatte ich auch schon, aber das dürfen wir wie gesagt leider nicht...

[c_newbie]

Bist schon weitergekommen? Hab das bei google gefunden

/*
 * 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.
 */

#define FALSE 0
#define TRUE 1

typedef struct element element;
struct element {
    element *next, *prev;
    int i;
};

int cmp(element *a, element *b) {
    return a->i - b->i;
}

/*
 * 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, int is_circular, int is_double) {
    element *p, *q, *e, *tail, *oldhead;
    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;
	oldhead = list;		       /* only used for circular linkage */
        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++;
		if (is_circular)
		    q = (q->next == oldhead ? NULL : q->next);
		else
		    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--;
		    if (is_circular && q == oldhead) q = NULL;
		} else if (qsize == 0 || !q) {
		    /* q is empty; e must come from p. */
		    e = p; p = p->next; psize--;
		    if (is_circular && p == oldhead) p = NULL;
		} else if (cmp(p,q) <= 0) {
		    /* First element of p is lower (or same);
		     * e must come from p. */
		    e = p; p = p->next; psize--;
		    if (is_circular && p == oldhead) p = NULL;
		} else {
		    /* First element of q is lower; e must come from q. */
		    e = q; q = q->next; qsize--;
		    if (is_circular && q == oldhead) q = NULL;
		}

                /* add the next element to the merged list */
		if (tail) {
		    tail->next = e;
		} else {
		    list = e;
		}
		if (is_double) {
		    /* Maintain reverse pointers in a doubly linked list. */
		    e->prev = tail;
		}
		tail = e;
            }

            /* now p has stepped `insize' places along, and q has too */
            p = q;
        }
	if (is_circular) {
	    tail->next = list;
	    if (is_double)
		list->prev = tail;
	} else
	    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;
    }
}

[c_newbie]

/*
 * 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.
 */

int main(void) {
    #define n 13
    element k[n], *head, *p;
    int is_circular, is_double;

    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 },
    };
    int i, j;

    for (j = 0; j < n; j++)
        k[j].i = j;

    listsort(NULL, 0, 0);

    for (is_circular = 0; is_circular < 2; is_circular++) {
	for (is_double = 0; is_double < 2; is_double++) {
	    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 = (is_circular ? &k[ord[0]] :
					  NULL);
		    else
			k[ord[j]].next = &k[ord[j+1]];
		    if (is_double) {
			if (j == 0)
			    k[ord[j]].prev = (is_circular ? &k[ord[n-1]] :
					      NULL);
			else
			    k[ord[j]].prev = &k[ord[j-1]];
		    }
		}

		printf("before:");
		p = head;
		do {
		    printf(" %d", p->i);
		    if (is_double) {
			if (p->next && p->next->prev != p)
			    printf(" [REVERSE LINK ERROR!]");
		    }
		    p = p->next;
		} while (is_circular ? (p != head) : (p != NULL));
		printf("\n");
		head = listsort(head, is_circular, is_double);
		printf(" after:");
		p = head;
		do {
		    printf(" %d", p->i);
		    if (is_double) {
			if (p->next && p->next->prev != p)
			    printf(" [REVERSE LINK ERROR!]");
		    }
		    p = p->next;
		} while (is_circular ? (p != head) : (p != NULL));
		printf("\n");
	    }
	}
    }
    return 0;
}

[Siderius]

ich habs jetzt selber... aber dennoch danke... hier meine lösung:

struct ListenElement * SortList (struct ListenElement * liste)
{
	struct ListenElement *ptr1, *ptr2, *temp1, *temp2;
	struct ListenElement TEMP;

	if(liste != NULL)					//ist in liste etwas drin?
	{
		ptr1 = liste;
		temp1 = liste;
		temp2 = liste;	

		while (ptr1 != NULL)					//solange in Liste etwas gefunden wird
		{
			ptr2 = ptr1->next;					//nächstes Element in Pointer 2
			while (ptr2 != NULL)				//Solange im nachfolgendem Inhalt ist
			{
				if (strcmp(ptr1->Name,ptr2->Name) > 0) //Zweiter Ausdruck > Erstem Ausdruck
				{
					temp2 = liste;				//Element in die "Zwischenablage"

					while(temp2->next != NULL)	//Solange Element nach TEMP2 Inhalt hat
					{
						if(temp2->next == ptr2) // Stimmt der Inhalt überein?
						{
							temp2->next = ptr1;	//Inhalt von Folge in den Zwischenspeicher
							break;
						}
						temp2 = temp2->next;	//Folge von Zwischenspeicher in den Vorgänger
					}

					if (ptr1 == liste)			//Wenn das erste Element = Liste
					{
						liste = ptr2;			//2. Nach vorne
					}
					else
					{
						temp1 = liste;		//element in ZWSP

						while(temp1->next != NULL)
						{
							if(temp1->next == ptr1)	//Stimmt Inhalt
							{
								temp1->next = ptr2;	//ZWSP
								break;
							}
							temp1 = temp1->next;	//Pointer um eine Position nach rechts
						}
					}

					//ÜBERSCHREIBEN DER DATEN
					TEMP.next = ptr1->next;
					ptr1->next = ptr2->next;
					ptr2->next = TEMP.next;

					ptr1 = ptr2;
					ptr2 = ptr1->next;
				}
				else
					ptr2 = ptr2->next;
			}
			ptr1 = ptr1->next;
		}
		return liste;

	}
	else
	{
		printf("SORT_ERROR!");
		return 0;
	}
}