我不知道如何删除c中链表中的第一个和最后一个元素。下面是我的程序，它使用链表。我不知道如何实际删除最后一个元素，但我能够找到它。该元素由一个整数和下一个指针组成。如果有人能帮助我，我将不胜感激。
struct ListNode{
int value;
struct ListNode *next;
};
typedef struct ListNode Link;
void deleteLast(Link *);
void printList(Link *);
void deleteFirst(Link *);
int main(){
Link *myList;
Link *curr, *newlink;
int i;
curr = myList;
for(i = 0; i < 10; i++){
newlink = (Link*) malloc(1*sizeof(Link));
newlink->value = i*i;
curr->next = newlink;
curr = newlink;
}
curr->next = NULL;
curr = myList->next;
deleteFirst(curr);
printList(curr);
printf("\n");
}
void deleteLast(Link *head)
{
Link *curr;
curr = head->next;
while (curr->next!=NULL)
{
curr = curr->next;
}
free(curr->next);
curr->next = NULL;
}
void printList(Link *head){
Link *curr;
curr = head->next;
printf("[");
if(curr!=NULL){
printf("%d",curr->value);
curr = curr->next;
}
while(curr != NULL){
printf(", %d", curr->value);
curr = curr->next;
}
printf("]\n");
}
void deleteFirst(Link *head){
Link *curr;
curr = head->next;
free(curr->value);
free(curr->next);
printf("%d\t",curr->value);
}
我尝试的都没有，请帮助我。
答案 0 :(得分：2)
您的代码中有很多错误。
curr->next = newlink;，其中curr = myList，初始化值。您可以将循环更改为Link *myList = NULL; Link *curr, *newlink; int i; curr = myList; for(i = 0; i < 10; i++){ newlink = malloc(1*sizeof(Link)); newlink->value = i*i; if (curr != NULL) curr->next = newlink; else myList = newlink; curr = newlink; }
Link *curr; curr = head->next; while (curr->next != NULL) { head = curr; curr = curr->next; } free(head->next); head->next = NULL;
Link** void deleteFirst(Link **head){ Link *curr; curr = (*head)->next; free(*head); *head = curr; }
您可以通过提供列表开头的地址从主要调用此函数：
deleteFirst(&myList);
deleteLast(myList);
printList(myList);
当然，在所有函数中，您必须检查列表中是否至少有一些值，而不是空指针NULL
答案 1 :(得分：0)
#include <stdint.h>
#include <stdlib.h>
#include <stdio.h>
typedef struct node Node;
struct node{
int element;
Node *next;
};
Node* head = NULL;
bool put(int data)
{
bool retVal = false;
Node* temp;
do
{
if (head == NULL)
{
head = malloc(sizeof(Node));
if (head)
{
printf("New head created\n");
head->element = data;
head->next = NULL;
printf("Element %d stored\n", data);
retVal = true;
break;
}
else
{
printf("Could not create new head\n");
break;
}
}
temp = head;
while (temp->next != NULL)
{
temp = temp->next;
}
temp->next = malloc(sizeof(Node));
temp->next->element = data;
temp->next->next = NULL;
printf("Element %d stored\n", data);
retVal = true;
break;
} while(1);
return retVal;
}
bool get(int *pData)
{
bool retVal = false;
Node* temp;
if (head != NULL)
{
retVal = true;
Node *prevNode = head;
temp = prevNode;
while (temp->next)
{
prevNode = temp;
temp = temp->next;
}
*pData = temp->element;
printf("Element %d retrieved\n", *pData);
free(temp);
printf("Node freed\n");
if (temp == head)
{
head = NULL;
}
else
{
prevNode->next = NULL;
}
}
return retVal;
}
int main(void)
{
int retrievedNum;
int num;
for (num = 0; num < 5; num++)
{
put(num);
}
printf("\n");
for (num = 0; num < 6; num++)
{
get(&retrievedNum);
}
put(num);
get(&retrievedNum);
printf("Element retrieved: %d\n", retrievedNum);
return 1;
}
答案 2 :(得分：0)
此代码适用于删除链接列表中的最后一个元素：
void dellast()
{
r=head;
struct node* z;
do
{
z=r;
r=r->next;
if(r->next==NULL)
{
z->next=NULL;
free(r->next);
}
}while(z->next!=NULL);
}
代码的工作原理如下：
保持当前节点及其上一个节点的跟踪。
如果current-＆gt; next == NULL表示它是最后一个节点。
请执行previous-＆gt; next = NULL并释放当前节点
答案 3 :(得分：0)
我在C中创建了我自己的（单个）LinkedList示例，它被证明有效：
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
/********** GLOBALS *******************************/
#define OK 0
#define ERROR -1
/********** STRUCT AND TYPES DEFINTIONS ***********/
/* a node with key, data and reference to next node*/
typedef struct Node {
int key;
char string[1024];
struct Node *next; // pointer to next node
} Node;
/* the actual linked list: ref to first and last Node, size attribute */
typedef struct LinkedList {
struct Node *first;
struct Node *last;
int size;
} LinkedList;
/********** FUNCTION HEADERS **********************/
LinkedList* init_list();
void insert_end(LinkedList *list, int key, char string[]);
void insert_beginning(LinkedList *list, int key, char string[]);
int remove_end(LinkedList *list);
int remove_beginning(LinkedList *list);
int print_list(LinkedList *list);
void free_list(LinkedList *list);
char * get_string(LinkedList *list, int key);
/*********** FUNCTION DEFINITIONS ***************/
/**
* init_list Returns an appropriately (for an empty list) initialized struct List
*
* @return LinkedList * ..ptr to the newly initialized list
*/
LinkedList * init_list() {
printf("initializing list...\n");
LinkedList *list = (LinkedList*) malloc(sizeof(LinkedList));
list->first = NULL;
list->last = NULL;
list->size = 0;
return list;
}
/**
* Given a List, a key and a string adds a Node containing this
* information at the end of the list
*
* @param list LinkedList * ..ptr to LinkedList
* @param key int .. key of the Node to be inserted
* @param string char[] .. the string of the Node to be inserted
*/
void insert_end(LinkedList *list, int key, char string[]) {
printf("----------------------\n");
list->size++; // increment size of list
// intialize the new Node
Node* newN = (Node*) malloc(sizeof(Node));
newN->key = key;
strcpy(newN->string, string);
newN->next = NULL;
Node* oldLast = list->last; // get the old last
oldLast->next = newN; // make new Node the next Node for oldlast
list->last = newN; // set the new last in the list
printf("new Node(%p) at end: %d '%s' %p \n", newN, newN->key, newN->string,newN->next);
}
/**
* Given a List, a key and a string adds a Node, containing
* this information at the beginning of the list
*
* @param list LinkedList * ..ptr to LinkedList
* @param key int .. key of the Node to be inserted
* @param string char[] .. the string of the Node to be inserted
*/
void insert_beginning(LinkedList *list, int key, char string[]) {
printf("----------------------\n");
list->size++; // increment size of list
Node* oldFirst = list->first; //get the old first node
/* intialize the new Node */
Node* newN = (Node*) malloc(sizeof(Node));
newN->key = key;
strcpy(newN->string, string);
newN->next = oldFirst;
list->first = newN; // set the new first
/* special case: if list size == 1, then this new one is also the last one */
if (list->size == 1)
list->last = newN;
printf("new Node(%p) at beginning: %d '%s' %p \n", newN, newN->key,newN->string, newN->next);
}
/**
* Removes the first Node from the list
*
* @param list LinkedList * .. ptr to the List
*
* @return OK | ERROR
*/
int remove_beginning(LinkedList *list) {
printf("----------------------\n");
if (list->size <= 0)
return ERROR;
list->size--;
Node * oldFirst = list->first;
printf("delete Node(%p) at beginning: '%d' '%s' '%p' \n", oldFirst,oldFirst->key, oldFirst->string, oldFirst->next);
free(list->first); //free it
list->first = oldFirst->next;
oldFirst = NULL;
return OK;
}
/**
* Removes the last Node from the list.
*
* @param list LinkedList * .. ptr to the List
*
* @return OK | ERROR
*/
int remove_end(LinkedList *list) {
printf("----------------------\n");
/* special case #1 */
if (list->size <= 0)
return ERROR;
/* special case #2 */
if (list->size == 1) {
free(list->first);
list->first = NULL;
list->last = NULL;
return OK;
}
printf("delete Node(%p) at end: '%d' '%s' '%p' \n", list->last,list->last->key, list->last->string, list->last->next);
list->size--; // decrement list size
Node * startNode = list->first;
/* find the new last node (the one before the old last one); list->size >= 2 at this point!*/
Node * newLast = startNode;
while (newLast->next->next != NULL) {
newLast = newLast->next;
}
free(newLast->next); //free it
newLast->next = NULL; //set to NULL to denote new end of list
list->last = newLast; // set the new list->last
return OK;
}
/**
* Given a List prints all key/string pairs contained in the list to
* the screen
*
* @param list LinkedList * .. ptr to the List
*
* @return OK | ERROR
*/
int print_list(LinkedList *list) {
printf("----------------------\n");
if (list->size <= 0)
return ERROR;
printf("List.size = %d \n", list->size);
Node *startN = list->first; //get first
/* iterate through list and print contents */
do {
printf("Node#%d.string = '%s', .next = '%p' \n", startN->key,startN->string, startN->next);
startN = startN->next;
} while (startN != NULL);
return OK;
}
/**
* Given a List, frees all memory associated with this list.
*
* @param list LinkedList * ..ptr to the list
*/
void free_list(LinkedList *list) {
printf("----------------------\n");
printf("freeing list...\n");
if (list != NULL && list->size > 0) {
Node * startN = list->first;
Node * temp = list->first;
do {
free(temp);
startN = startN->next;
temp = startN;
} while (startN != NULL);
free(list);
}
}
/**
* Given a List and a key, iterates through the whole List and returns
* the string of the first node which contains the key
*
* @param list LinkedList * ..ptr to the list
* @param key int .. the key of the Node to get the String from
*
* @return OK | ERROR
*/
char * get_string(LinkedList *list, int key) {
printf("----------------------\n");
Node *startN = list->first; //get first
/* iterate through list and find Node where node->key == key */
while (startN->next != NULL) {
if (startN->key == key)
return startN->string;
else
startN = startN->next;
}
return NULL;
}
/*************** MAIN **************/
int main(void) {
LinkedList *list = init_list();
insert_beginning(list, 1, "im the first");
insert_end(list, 2, "im the second");
insert_end(list, 3, "im the third");
insert_end(list, 4, "forth here");
print_list(list);
remove_end(list);
print_list(list);
remove_beginning(list);
print_list(list);
remove_end(list);
print_list(list);
printf("string at node with key %d = '%s' \n",2,get_string(list, 2));
free_list(list);
return OK;
}
希望这个实现示例有所帮助。