如何从链表中删除节点?
这是我的代码:
void RemoveNode(Node * node, Node ** head) {
if (strcmp(node->state, (*(*head)->next).state) == 0) {
Node * temp = *head;
*head = (*head)->next;
free(temp);
return;
}
Node * current = (*head)->next;
Node * previous = *head;
while (current != NULL && previous != NULL) {
if (strcmp(node->state, (*current->next).state) == 0) {
Node * temp = current;
previous->next = current->next;
free(temp);
return;
}
current = current->next;
previous = previous->next;
}
return;
}
但我总是犯seg错误。
我觉得我在做一些愚蠢的事情。。。。有什么想法吗?
我的猜测:
void RemoveNode(Node * node, Node ** head) {
if (strcmp(node->state, ((*head)->state) == 0) {
Node * temp = *head;
*head = (*head)->next;
free(temp);
return;
}
Node * current = (*head)->next;
Node * previous = *head;
while (current != NULL && previous != NULL) {
if (strcmp(node->state, current->state) == 0) {
Node * temp = current;
previous->next = current->next;
free(temp);
return;
}
previous = current;
current = current->next;
}
return;
}
我建议您尝试使用递归来执行此操作,以避免使用"双指针"。它将极大地简化逻辑。这个链接有一个很好的解释和递归实现。如果您试图从空链接列表中删除一个节点,此方法也会特别有效。
Node *ListDelete(Node *currP, State value)
{
/* See if we are at end of list. */
if (currP == NULL)
return NULL;
/*
* Check to see if current node is one
* to be deleted.
*/
if (currP->state == value) {
Node *tempNextP;
/* Save the next pointer in the node. */
tempNextP = currP->next;
/* Deallocate the node. */
free(currP);
/*
* Return the NEW pointer to where we
* were called from. I.e., the pointer
* the previous call will use to "skip
* over" the removed node.
*/
return tempNextP;
}
/*
* -------------- RECURSION-------------------
* Check the rest of the list, fixing the next
* pointer in case the next node is the one
* removed.
*/
currP->next = ListDelete(currP->next, value);
/*
* Return the pointer to where we were called
* from. Since we did not remove this node it
* will be the same.
*/
return currP;
}