该程序创建一个在开头插入的链表,并具有一个 deleteall(key)
函数,该函数将删除具有该键值的所有节点
#include <stdio.h>
#include <stdlib.h>
struct node{
int data;
struct node* next; //creation a linked list with insertion at beginning
};
struct node* head = NULL;
int totaldeleted = 0;
void create(n) {
int i;
struct node *newnode, *temp;
newnode = (struct node*)malloc(sizeof(struct node));
printf("Enter the data: ");
scanf("%d", &newnode -> data);
newnode -> next = NULL;
if(head==NULL){
head=newnode;
}
temp = head;
for(i=2; i<=n; i++){
int data;
newnode = (struct node*)malloc(sizeof(struct node));
printf("Enter the data: ");
scanf("%d", &data);
newnode -> data = data;
newnode -> next = NULL;
newnode -> next = head;
head = newnode;
}
}
void display(){
struct node* temp;
printf("\nThe Linked list is: ");
temp = head;
while(temp!= NULL){
printf("%d", temp -> data);
temp = temp -> next;
}
}
int deleteall(key){
struct node *prev, *cur;
if(head == NULL){
prev = NULL;
cur = NULL;
printf("\nList is empty!");
}
while(head != NULL && head->data == key){
prev = head;
head = head -> next;
free(prev);
totaldeleted++;
}
prev = NULL;
cur = head;
while(cur != NULL){
if(cur->data == key && prev != NULL){
prev -> next = cur->next;
free(cur);
cur = prev->next; // I cannot understand this logic.
totaldeleted++;
}
else {
prev = cur;
cur = cur -> next;
}
}
return totaldeleted;
}
int main(){
int n, key;
printf("Enter the number of nodes: ");
scanf("%d", &n);
create(n);
display();
printf("\nEnter the key: ");
scanf("%d", &key);
totaldeleted = deleteall(key);
printf("Total deleted: %d", totaldeleted);
display();
printf("\n");
}
最佳答案
就在该行之前,cur
指向已删除节点 - 该节点现已释放,因此 cur
指针的值不确定。我们希望将 cur
指向已删除节点后面的节点。如果它没有被删除,我们可以从cur->next
获取它。但是,cur->next
无效。我们可以在 free
...
cur->next
的值
但是,在free
之前有这样一行:
prev -> next = cur->next;
即现在我们需要的值已经分配到了 prev->next
中,因此我们不需要临时变量,但可以从 prev->next
中获取它。
关于c - 我不明白deleteall(key)函数中这条语句 "cur = prev->next;"的逻辑,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/51579890/