[英]Deleting front node from a Singly Linked List in C
我試圖從單鏈接列表中刪除,但是,當我嘗試從第一個元素中刪除時,它會打印垃圾。 我認為問題來自delete_node函數,但是,我嘗試了所有操作,但無法弄清楚。
#include <stdio.h>//prinf
#include <stdlib.h>//alloc mallco callo
typedef struct node node;
struct node{
int number;
node *next;
};
node *new_node(int num){
node *n= (node*) malloc(sizeof(node));
n->number=num;
n->next=NULL;
return n;
}
void node_free_all(node *n){
if(n != NULL){
node_free_all(n->next);
free(n);
}
}
void print_nodes(node *n){
if(n != NULL){
print_nodes(n->next);
printf("Number is: %d\n",n->number);
}
}
void delete_node(node *n, int num){
node *rmNode= (node*)malloc(sizeof(node));
//delete first
if( n!= NULL && n->number==num){
rmNode = n;
n=n->next;
free(rmNode);
}
//all but first
while(n != NULL){
if(n->next != NULL && n->next->number == num){
rmNode= n->next;
n->next = rmNode->next;
free(rmNode);
break;
}
n=n->next;
}
}
int main(){
int i;
node *head= (node*) malloc(sizeof(node));
node *curr;
head=NULL;
for(i=1;i<=10;i++) {
curr = new_node(i);
curr->next =head;
head=curr;
}
printf("Everything:\n");
print_nodes(head);
printf("Deleting 1:\n");
delete_node(head,1);
print_nodes(head);
printf("Deleting 5:\n");
delete_node(head,5);
print_nodes(head);
printf("Deleting 2:\n");
delete_node(head,2);
print_nodes(head);
printf("Deleting 3:\n");
delete_node(head,3);
print_nodes(head);
printf("Deleting 10:\n");
delete_node(head,10);
print_nodes(head);
printf("Deleting 9:\n");
delete_node(head,9);
print_nodes(head);
node_free_all(head);
// node_free_all(list);
return 0;
}
我究竟做錯了什么?
您不需要為rmNode
分配內存。 另外,您還需要將頭指針的引用傳遞給delete_node
函數,因為每次更新列表時,如果必須刪除列表的第一個元素,則在這種情況下,頭指針也會被更新。
struct node
{
int number;
node *next;
};
void delete_node(struct node** head_ref, int num)
{
struct node* temp;
struct node* current = (*head_ref);
//delete first
if( current != NULL && current->number == num)
{
temp = current;
current = current->next;
free(temp);
(*head_ref) = current;
}
else
{
//all but first
while(current != NULL)
{
if(current->next != NULL && current->next->number == num)
{
temp = current->next;
current->next = temp->next;
free(temp);
break;
}
current = current->next;
}
}
}
這是一個版本,該版本會刪除所有與num
匹配的項目,並始終更新主項目。
void delete_node(node** head_ref, int num)
{
node* temp;
node* last = 0;
node* current = *head_ref;
while(current)
{
if( current->number == num )
{
temp = current;
if( current == *head_ref )
current = (*head_ref) = current->next;
else
current = last->next = current->next;
free(temp);
} else {
last = current;
current = current->next;
}
}
}
您必須像delete_node(&head,1);
這樣稱呼它delete_node(&head,1);
因為它需要一個可以更改標題的地址
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