我正在编写一个程序,它创建一个双向链表并从中删除某个元素。一切都很有效,除了列表只包含1个元素的部分,当我尝试删除它时,程序崩溃。有什么建议吗?
#include <stdio.h>
#include <stdlib.h>
struct Node{
int data;
struct Node* next;
struct Node* prev;
};
struct Node* head;
struct Node* tail;
struct Node* CreateNewNode(int x){
struct Node* newNode=(struct Node*)malloc(sizeof(struct Node));
newNode->data=x;
newNode->next=NULL;
newNode->prev=NULL;
return newNode;
}
void InsertAtBegin(int x){
struct Node* newNode=CreateNewNode(x);
if(head == NULL) {
head = newNode;
tail = newNode;
return;
}
head->prev = newNode;
newNode->next = head;
head = newNode;
}
void InsertAtEnd(int x){
struct Node* newNode=CreateNewNode(x);
if(tail==NULL){
head=newNode;
tail=newNode;
return;
}
tail->next=newNode;
newNode->prev=tail;
tail=newNode;
}
struct Node* PointTo(int k){
struct Node* curr=head;
int i=1;
while(curr!=NULL && i<k){
curr=curr->next;
i++;
}
return curr;
}
void DelFromList(int k){
struct Node* temp;
temp=PointTo(k);
if(k==1){
head=temp->next;
temp->next->prev=NULL;
free(temp);
return;
}
if(temp->next==NULL){
temp->prev->next=NULL;
tail=temp->prev;
free(temp);
return;
}
if(temp->next==NULL && temp->prev==NULL){
head=NULL;
tail=NULL;
printf("atpazista\n");
free(temp);
return;
}
temp->next->prev=temp->prev;
temp->prev->next=temp->next;
free(temp);
}
void Print(){
struct Node* temp = head;
if(head==NULL){
printf("List empty\n");
return;
}
printf("List: ");
while(temp != NULL) {
printf("%d ",temp->data);
temp = temp->next;
}
printf("\n");
}
void Read(int *x){
while(scanf("%d", x)==0){
printf("NUMBERS ONLY\n");
scanf("%s");
}
}
void Free(){
struct Node* temp=head;
while(temp->next!=NULL){
temp=temp->next;
free(temp->prev);
}
}
int main()
{
int n=0, x;
printf("Number of elements?\n");
while(n<=0){
Read(&n);
}
int i;
for(i=0; i<n; i++){
printf("Type a number\n");
Read(&x);
InsertAtEnd(x);
}
Print();
printf("Head: %d\n", head->data);
printf("Tail: %d\n", tail->data);
printf("Number of element to be deleted?\n");
n=0;
while(n<=0){
Read(&n);
}
DelFromList(n);
Print();
printf("Head: %d\n", head->data);
printf("Tail: %d\n", tail->data);
Free();
return 0;
}
答案 0 :(得分:2)
也许交换了以下的顺序:
if(temp->next==NULL){
temp->prev->next=NULL;
tail=temp->prev;
free(temp);
return;
}
if(temp->next==NULL && temp->prev==NULL){
head=NULL;
tail=NULL;
printf("atpazista\n");
free(temp);
return;
}
后一个代码永远不会执行,因为第一个案例已经执行,并返回