[英]inserting a node into a linked list
Assume structure list and node are defined as假设结构列表和节点定义为
struct list {struct node *a;};
struct node { int value;
struct node *next;};
The following function inserts integer e into l as the first element以下函数将整数 e 作为第一个元素插入 l
void insert_first(int e, struct list *l){
struct node * r = malloc(sizeof(struct node));
r->value = e;
r->next = l->a;
l->a = r;}
Example: original list "b": 1 2 3 4示例:原始列表“b”:1 2 3 4
after calling insert_first(3,*b)调用 insert_first(3,*b) 后
list "b": 3 1 2 3 4列表“b”:3 1 2 3 4
insert_first is pretty straightfoward; insert_first 非常简单; however, I am having a hard time trying to figure out how to write a function insert_last which inserts a number as the last element of the list.
但是,我很难弄清楚如何编写一个函数 insert_last 来插入一个数字作为列表的最后一个元素。
Example: original list "b": 1 2 3 4示例:原始列表“b”:1 2 3 4
after calling insert_last(3,*b)调用 insert_last(3,*b) 后
list "b": 1 2 3 4 3列表“b”:1 2 3 4 3
Thanks for any help in advance.提前感谢您的任何帮助。
You need to save original HEAD node and traverse through list .您需要保存原始 HEAD 节点并遍历 list 。 Hope this code will help you.
希望这段代码能帮到你。
struct node {
int value;
struct node *next;
};
struct list {struct node *a;};
struct node *insert_last(int e, struct list *l) {
/* Store the initial head of the list */
struct list *org_head = head;
struct node *r = malloc(sizeof(struct node));
r->value = e;
r->next = NULL /* Assign next pointer of current node to NULL */
/* If the head is initially NULL, then directly return the new created node (r) as the head of a linked list with only one node */
if(head == NULL)
{
return r;
}
/* While we do not reach the last node, move to the next node */
while(head -> next != NULL)
head = head -> next;
/* Assign the 'next' pointer of the current node to the "r" */
head->next = r;
/* return the original head that we have stored separately before */
return org_head;
}
One way of doing it is to iterate over the list until you find the tail.一种方法是遍历列表直到找到尾部。 Something like this:
像这样的东西:
void insert_last(int e, struct list *l)
{
// "iter" Will iterate over the list.
struct node *iter = l->a;
struct node *new_node = malloc(sizeof(struct node));
// Advice: ALWAYS check, if malloc returned a pointer!
if(!new_node) exit(1); // Memory allocation failure.
new_node->value = e;
new_node->next = NULL;
if(iter){
// Loop until we found the tail.
// (The node with no next node)
while(iter->next) iter = iter->next;
// Assign the new tail.
iter->next = new_node;
}else{
// The list was empty, assign the new node to be the head of the list.
l->a = new_node;
}
}
EDIT: Something I saw in your code, that really tickles me: ALWAYS check, when using malloc, whether you actually got a pointer back or not (check if the pointer is NULL).编辑:我在你的代码中看到的东西,真的让我很痒:在使用 malloc 时,总是检查你是否真的得到了一个指针(检查指针是否为 NULL)。 If malloc failes to allocate memory, be it for a lack thereof or some other critical error, it will toss you a NULL pointer.
如果 malloc 无法分配内存,无论是由于缺少内存还是其他一些严重错误,它都会向您抛出一个 NULL 指针。 If you do not check for that, you might end up running in to some very nasty, hard to detect bugs.
如果您不检查这一点,您最终可能会遇到一些非常讨厌、难以检测的错误。 Just a little reminder!
只是一点提醒!
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