我在C中使用链接列表的插入方法遇到了一些问题。它似乎只在列表的开头添加。我做的任何其他插入都失败了。而这个CodeBlocks调试器很难理解我仍然没有得到它。它永远不会给我价值,只有内存中的地址。无论如何这是我的功能。你有没有看到它失败的原因?
/* function to add a new node at the end of the list */
int addNodeBottom(int val, node *head){
//create new node
node *newNode = (node*)malloc(sizeof(node));
if(newNode == NULL){
fprintf(stderr, "Unable to allocate memory for new node\n");
exit(-1);
}
newNode->value = val;
//check for first insertion
if(head->next == NULL){
head->next = newNode;
printf("added at beginning\n");
}
else
{
//else loop through the list and find the last
//node, insert next to it
node *current = head;
while(current->next != NULL)
{
if(current->next == NULL)
{
current->next = newNode;
printf("added later\n");
}
current = current->next;
}
}
return 0;
}
然后在main中,只添加了929。
//testing addNodeBottom function
addNodeBottom(929, head);
addNodeBottom(98, head);
addNodeBottom(122, head);
addNodeBottom(11, head);
addNodeBottom(1034, head);
答案 0 :(得分:12)
此代码可以使用。 samplebias的答案几乎是正确的,但您需要进行第三次更改:
int addNodeBottom(int val, node *head){
//create new node
node *newNode = (node*)malloc(sizeof(node));
if(newNode == NULL){
fprintf(stderr, "Unable to allocate memory for new node\n");
exit(-1);
}
newNode->value = val;
newNode->next = NULL; // Change 1
//check for first insertion
if(head->next == NULL){
head->next = newNode;
printf("added at beginning\n");
}
else
{
//else loop through the list and find the last
//node, insert next to it
node *current = head;
while (true) { // Change 2
if(current->next == NULL)
{
current->next = newNode;
printf("added later\n");
break; // Change 3
}
current = current->next;
};
}
return 0;
}
更改1:newNode->next
必须设置为NULL
,因此我们不会在列表末尾插入无效指针。
更改2/3:循环更改为无限循环,当我们找到最后一个元素时,将使用break;
跳出。请注意while(current->next != NULL)
之前如何与if(current->next == NULL)
相矛盾。
编辑:关于while循环,这种方式更好:
node *current = head;
while (current->next != NULL) {
current = current->next;
}
current->next = newNode;
printf("added later\n");
答案 1 :(得分:2)
在您malloc
后node
确保设置node->next = NULL
。
int addNodeBottom(int val, node *head)
{
node *current = head;
node *newNode = (node *) malloc(sizeof(node));
if (newNode == NULL) {
printf("malloc failed\n");
exit(-1);
}
newNode->value = val;
newNode->next = NULL;
while (current->next) {
current = current->next;
}
current->next = newNode;
return 0;
}
我应该指出,在这个版本中,head
仍然用作虚拟,不用于存储值。这使您可以通过仅head
节点来表示空列表。
答案 2 :(得分:0)
我想在编写代码之前提及密钥供您考虑。
//键
temp = malloc函数分配的新节点的地址(C中的成员od alloc.h库)
prev =现有链接列表的最后一个节点的地址。
next =包含下一个节点的地址
struct node {
int data;
struct node *next;
} *head;
void addnode_end(int a) {
struct node *temp, *prev;
temp = (struct node*) malloc(sizeof(node));
if (temp == NULL) {
cout << "Not enough memory";
} else {
node->data = a;
node->next = NULL;
prev = head;
while (prev->next != NULL) {
prev = prev->next;
}
prev->next = temp;
}
}
答案 3 :(得分:0)
始终在给定链接列表的最后一个节点之后添加新节点。例如,如果给定的链接列表是5-> 10-> 15-> 20-> 25并且我们在末尾添加项目30,则链接列表变为5-> 10-> 15- &GT; 20→; 25-大于30。 由于链接列表通常由其头部表示,因此我们必须遍历列表直到结束,然后将最后一个节点更改为新节点。
/* Given a reference (pointer to pointer) to the head
of a list and an int, appends a new node at the end */
void append(struct node** head_ref, int new_data)
{
/* 1. allocate node */
struct node* new_node = (struct node*) malloc(sizeof(struct node));
struct node *last = *head_ref; /* used in step 5*/
/* 2. put in the data */
new_node->data = new_data;
/* 3. This new node is going to be the last node, so make next
of it as NULL*/
new_node->next = NULL;
/* 4. If the <a href="#">Linked List</a> is empty, then make the new node as head */
if (*head_ref == NULL)
{
*head_ref = new_node;
return;
}
/* 5. Else traverse till the last node */
while (last->next != NULL)
last = last->next;
/* 6. Change the next of last node */
last->next = new_node;
return;
}
答案 4 :(得分:0)
这很好用:
struct node *addNode(node *head, int value) {
node *newNode = (node *) malloc(sizeof(node));
newNode->value = value;
newNode->next = NULL;
if (head == NULL) {
// Add at the beginning
head = newNode;
} else {
node *current = head;
while (current->next != NULL) {
current = current->next;
};
// Add at the end
current->next = newNode;
}
return head;
}
使用示例:
struct node *head = NULL;
for (int currentIndex = 1; currentIndex < 10; currentIndex++) {
head = addNode(head, currentIndex);
}
答案 5 :(得分:0)
我知道这是一篇旧文章,仅供参考。这是在没有特殊情况检查的情况下添加空列表的方法,尽管以看起来更复杂的代码为代价。
void Append(List * l, Node * n)
{
Node ** next = &list->Head;
while (*next != NULL) next = &(*next)->Next;
*next = n;
n->Next = NULL;
}