[英]Why are we using pointer to pointer for one and a normal pointer to another?
// A complete working C++ program to demonstrate
// all insertion methods on Linked List
#include <bits/stdc++.h>
using namespace std;
// A linked list node
class Node
{
public:
int data;
Node *next;
};
/* Given a reference (pointer to pointer)
to the head of a list and an int, inserts
a new node on the front of the list. */
void push(Node** head_ref, int new_data)
{
/* 1. allocate node */
Node* new_node = new Node();
/* 2. put in the data */
new_node->data = new_data;
/* 3. Make next of new node as head */
new_node->next = (*head_ref);
/* 4. move the head to point to the new node */
(*head_ref) = new_node;
}
/* Given a node prev_node, insert a new node after the given
prev_node */
void insertAfter(Node* prev_node, int new_data)
{
/*1. check if the given prev_node is NULL */
if (prev_node == NULL)
{
cout<<"The given previous node cannot be NULL";
return;
}
/* 2. allocate new node */
Node* new_node = new Node();
/* 3. put in the data */
new_node->data = new_data;
/* 4. Make next of new node as next of prev_node */
new_node->next = prev_node->next;
/* 5. move the next of prev_node as new_node */
prev_node->next = new_node;
}
/* Given a reference (pointer to pointer) to the head
of a list and an int, appends a new node at the end */
void append(Node** head_ref, int new_data)
{
/* 1. allocate node */
Node* new_node = new Node();
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 Linked List 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;
}
// This function prints contents of
// linked list starting from head
void printList(Node *node)
{
while (node != NULL)
{
cout<<" "<<node->data;
node = node->next;
}
}
/* Driver code*/
int main()
{
/* Start with the empty list */
Node* head = NULL;
// Insert 6. So linked list becomes 6->NULL
append(&head, 6);
// Insert 7 at the beginning.
// So linked list becomes 7->6->NULL
push(&head, 7);
// Insert 1 at the beginning.
// So linked list becomes 1->7->6->NULL
push(&head, 1);
// Insert 4 at the end. So
// linked list becomes 1->7->6->4->NULL
append(&head, 4);
// Insert 8, after 7. So linked
// list becomes 1->7->8->6->4->NULL
insertAfter(head->next, 8);
cout<<"Created Linked list is: ";
printList(head);
return 0;
}
// from geeksforgeek
为什么 append 和 push 函数有一个指向指针的指针(head_ref),而 insertAfter 函数有一个普通的指针(prev_node)? 由于我们正在更改它们是头节点还是前一个节点,因此应该是相同的方法(指针类型)。
我可以使用普通指针而不是指向指针的指针吗?
函数push
和append
都需要能够改变指向头节点的指针。
如果他们只传递了一个指向头Node*
,那么他们将只能将该指针的本地副本更改为头节点。 但是,它们需要能够更改指向头节点的原始指针,即函数main
的变量head
。 这就是为什么他们需要一个指向函数main
变量head
的指针,所以两个函数都需要传递一个Node**
。
与push
和append
函数insertAfter
,函数insertAfter
永远不需要更改传递给函数的指针。 这就是为什么传递一个Node*
就足够了。
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