从链接列表中删除项目(Remove an Item from Linked List)
实现 (Implementation)
该算法的实现如下 -
#include <stdio.h>
#include <stdlib.h>
struct node {
int data;
struct node *prev;
struct node *next;
};
struct node *head = NULL;
struct node *last = NULL;
struct node *current = NULL;
//display the list
void printList() {
struct node *ptr = head;
printf("\n[head] <=>");
//start from the beginning
while(ptr != NULL) {
printf(" %d <=>",ptr->data);
ptr = ptr->next;
}
printf(" [last]\n");
}
void print_backward() {
struct node *ptr = last;
printf("\n[last] <=>");
//start from the beginning
while(ptr != NULL) {
printf(" %d <=>",ptr->data);
ptr = ptr->prev;
}
printf(" [head]\n");
}
//Create Linked List
void insert(int data) {
// Allocate memory for new node;
struct node *link = (struct node*) malloc(sizeof(struct node));
link->data = data;
link->prev = NULL;
link->next = NULL;
// If head is empty, create new list
if(head==NULL) {
head = link;
return;
}
current = head;
// move to the end of the list
while(current->next!=NULL)
current = current->next;
// Insert link at the end of the list
current->next = link;
last = link;
link->prev = current;
}
void remove_data(int data) {
int pos = 0;
struct node *pre_node;
if(head==NULL) {
printf("Linked List not initialized");
return;
}
if(head->data == data) {
if(head->next != NULL) {
head->next->prev = NULL;
head = head->next;
return;
} else {
head = NULL;
printf("List is empty now");
return;
}
} else if(head->data != data && head->next == NULL) {
printf("%d not found in the list\n", data);
return;
}
current = head;
while(current->next != NULL && current->data != data) {
pre_node = current;
current = current->next;
}
if(current->data == data) {
pre_node->next = pre_node->next->next;
if(pre_node->next != NULL) { // link back
pre_node->next->prev = pre_node;
} else
last = pre_node;
free(current);
} else
printf("%d not found in the list.", data);
}
int main() {
insert(10);
insert(20);
insert(30);
insert(1);
insert(40);
insert(56);
printList();
remove_data(20);
printList();
print_backward();
return 0;
}
输出 (Output)
该方案的产出应该是 -
[head] <=> 10 <=> 20 <=> 30 <=> 1 <=> 40 <=> 56 <=> [last]
[head] <=> 10 <=> 30 <=> 1 <=> 40 <=> 56 <=> [last]
[last] <=> 56 <=> 40 <=> 1 <=> 30 <=> 10 <=> [head]
