标签：malloc type tail lca and 删除节点 value 缺点 构建
1、遍历链表
先将head指针赋值给一个局部变量current：
//return the number of nodes in a list (while-loop version)
int Length(struct node* head)
{
int count = 0;
struct node* current = head;
while (current != NULL)
{
count++;
current = current->next;
}
return count;
}当然也可以写为：
for (current = head; current != NULL; current = current->next) {}2、通过传递reference pointer改变某个指针
看个例子：
//Change the passed in head pointer to be NULL
//Uses a reference pointer to access the caller's memory
void ChangeToNull(struct node** headRef) //takes a pointer to the value of interest
{
*headRef = NULL; //use * to access the value of interest
}
void ChangeCaller()
{
struct node* head1;
struct node* head2;
ChangeToNull(&head1); //use & to compute and pass a pointer to
ChangeToNull(&head2); //the value of interest
//head1 and head2 are NULL at this point
}这块的思想是和（一）中的Push()类似。
内存示意图：
3、通过Push()建立链表（头插法）
这种方式的优点是速度飞快，简单易行，缺点是得到的链表是逆序的：
struct node* AddAtHead()
{
struct node* head = NULL;
for (int i = 1; i < 6; i++)
{
Push(&head, i);
}
//head == {5,4,3,2,1};
return head;
}4、尾插法建立链表
这种方法需要找到链表最后一个节点，改变其.next域：
.next域；head指针。5、特例+尾插法
如果要构建一个新的链表，那么头节点就要单独处理：
struct node* BuildWithSpecialCase()
{
struct node* head = NULL;
struct node* tail;
//deal with the head node here, and set the tail pointer
Push(&head, 1);
tail = head;
//do all the other nodes using "tail"
for (int i = 2; i < 6; i++)
{
Push(&(tail->next), i); //add node at tail->next
tail = tail->next; //advance tail to point to last node
}
return head; //head == {1,2,3,4,5}
}6、临时节点建立
struct node* BuildWithDummyNode()
{
struct node dummy; //dummy node is temporarily the first node
struct node* tail = &dummy; //build the list on dummy.next
dummy.next = NULL;
for (int i = 1; i < 6; i++)
{
Push(&(tail->next), i);
tail = tail->next;
}
//the real result list is now in dummy.next
//dummy.next == {1,2,3,4,5}
return dummy.next;
}7、本地指针建立
struct node* BuildWithLocalRef()
{
struct node* head = NULL;
struct node** lastPtrRef = &head; //start out pointing to the head pointer
for (int i = 1; i < 6; i++)
{
Push(lastPtrRef, i); //add node at the last pointer in the list
//advance to point to the new last pointer
lastPtrRef = &((*lastPtrRef)->next);
}
return head; //head == {1,2,3,4,5}
}这块可能有些抽象：
1）lastPtrRef开始指向head指针，以后指向链表最后一个节点中的.next域；
2）在最后加上一个节点；
3）让lastPtrRef指针向后移动，指向最后一个节点的.next域。 (*lastPtrRef)->next可以理解为*lastPtrRef指针指向的节点的next域。
1、AppendNode()
1) 不使用Push()函数：
struct node* AppendNode(struct node** headRef, int num)
{
struct node* current = *headRef;
struct node* newNode;
newNode = (struct node*)malloc(sizeof(struct node));
newNode->data = num;
newNode->next = NULL;
//special case for length 0
if (current == NULL)
{
*headRef = newNode;
}
else
{
//Locate the last node
while (current->next != NULL)
{
current = current->next;
}
current->next = newNode;
}
}2) 使用Push()函数：
struct node* AppendNode(struct node** headRef, int num)
{
struct node* current = *headRef;
//special case for length 0
if (current == NULL)
{
Push(headRef, num);
}
else
{
//Locate the last node
while (current->next != NULL)
{
current = current->next;
}
//Build the node after the last node
Push(&(current->next), num);
}
}2、CopyList
用一个指针遍历原来的链表，两个指针跟踪新生成的链表（一个head，一个tail）。
1) 不使用Push()函数：
struct node* CopyList(struct node* head)
{
struct node* current = head; //used to iterate over the original list
struct node* newList = NULL; //head of the new list
struct node* tail = NULL; //kept pointing to the last node in the new list
while (current != NULL)
{
if (newList == NULL) //special case for the first new node
{
newList = (struct node*)malloc(sizeof(struct node));
newList->data = current->data;
newList->next = NULL;
tail = newList;
}
else
{
tail->next = (struct node*)malloc(sizeof(struct node));
tail = tail->next;
tail->data = current->data;
tail->next = NULL;
}
current = current->next;
}
return newList;
}内存示意图：
2) 使用Push()函数：
struct node* CopyList2(struct node* head)
{
struct node* current = head; //used to iterate over the original list
struct node* newList = NULL; //head of the new list
struct node* tail = NULL; //kept pointing to the last node in the new list
while (current != NULL)
{
if (newList == NULL) //special case for the first new node
{
Push(&newList, current->data);
tail = newList;
}
else
{
Push(&(tail->next), current->data); //add each node at the tail
tail = tail->next; //advance the tail to the new last node;
}
current = current->next;
}
return newList;
}3) 使用Dummy Node：
struct node* CopyList3(struct node* head)
{
struct node* current = head; //used to iterate over the original list
struct node* tail = NULL; //kept pointing to the last node in the new list
struct node dummy; //build the new list off this dummy node
dummy.next = NULL;
tail = &dummy; //start the tail pointing at the dummy
while (current != NULL)
{
Push(&(tail->next), current->data); //add each node at the tail
tail = tail->next; //advance the tail to the new last node
current = current->next;
}
return dummy.next;
}4) 使用Local References：
struct node* CopyList4(struct node* head)
{
struct node* current = head; //used to iterate over the original list
struct node* newList = NULL; //head of the new list
struct node** lastPtr;
lastPtr = &newList; //start off pointing to the head itself
while (current != NULL)
{
Push(lastPtr, current->data); //add each node at the lastPtr
lastPtr = &((*lastPtr)->next); //advance lastPtr
current = current->next;
}
return newList;
}核心思想是使用lastPtr指针指向每个节点的.next域这个指针，而不是指向节点本身。
Recursive：struct node* CopyList5(struct node* head)
{
struct node* current = head;
if (head == NULL)
return NULL;
else {
struct node* newList = (struct node*)malloc(sizeof(struct node)); //make one node
newList->data = current->data;
newList->next = CopyList5(current->next); //recur for the rest
return newList;
}
}标签：malloc type tail lca and 删除节点 value 缺点 构建
原文地址：https://www.cnblogs.com/EIMadrigal/p/12130892.html
