原文：
给定一个链表，使用递归函数反向打印。 例如，如果给定的链表为1 -> 2 -> 3 -> 4，则输出应为4 -> 3 -> 2 -> 1。
注意，问题仅在于打印反面。 要反转列表本身，请参阅。
难度级别：菜鸟
算法
printreverse(head)
1\. call print reverse for head->next
2\. print head->data
实现：
c
// c program to print reverse of a linked list
#include
using namespace std;
/* link list node */
class node
{
public:
int data;
node* next;
};
/* function to reverse the linked list */
void printreverse(node* head)
{
// base case
if (head == null)
return;
// print the list after head node
printreverse(head->next);
// after everything else is printed, print head
cout << head->data << " ";
}
/*utility functions*/
/* push a node to linked list. note that this function
changes the head */
void push(node** head_ref, char new_data)
{
/* allocate node */
node* new_node = new node();
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to pochar to the new node */
(*head_ref) = new_node;
}
/* driver code*/
int main()
{
// let us create linked list 1->2->3->4
node* head = null;
push(&head, 4);
push(&head, 3);
push(&head, 2);
push(&head, 1);
printreverse(head);
return 0;
}
// this code is contributed by rathbhupendra
c
// c program to print reverse of a linked list
#include
#include
/* link list node */
struct node
{
int data;
struct node* next;
};
/* function to reverse the linked list */
void printreverse(struct node* head)
{
// base case
if (head == null)
return;
// print the list after head node
printreverse(head->next);
// after everything else is printed, print head
printf("%d ", head->data);
}
/*utility functions*/
/* push a node to linked list. note that this function
changes the head */
void push(struct node** head_ref, char new_data)
{
/* allocate node */
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to pochar to the new node */
(*head_ref) = new_node;
}
/* driver program to test above function*/
int main()
{
// let us create linked list 1->2->3->4
struct node* head = null;
push(&head, 4);
push(&head, 3);
push(&head, 2);
push(&head, 1);
printreverse(head);
return 0;
}
java
// java program to print reverse of a linked list
class linkedlist
{
node head; // head of list
/* linked list node*/
class node
{
int data;
node next;
node(int d) {data = d; next = null; }
}
/* function to print reverse of linked list */
void printreverse(node head)
{
if (head == null) return;
// print list of head node
printreverse(head.next);
// after everything else is printed
system.out.print(head.data " ");
}
/* utility functions */
/* inserts a new node at front of the list. */
public void push(int new_data)
{
/* 1 & 2: allocate the node &
put in the data*/
node new_node = new node(new_data);
/* 3\. make next of new node as head */
new_node.next = head;
/* 4\. move the head to point to new node */
head = new_node;
}
/*driver function to test the above methods*/
public static void main(string args[])
{
// let us create linked list 1->2->3->4
linkedlist llist = new linkedlist();
llist.push(4);
llist.push(3);
llist.push(2);
llist.push(1);
llist.printreverse(llist.head);
}
}
/* this code is contributed by rajat mishra */
python3
# python3 program to print reverse
# of a linked list
# node class
class node:
# constructor to initialize
# the node object
def __init__(self, data):
self.data = data
self.next = none
class linkedlist:
# function to initialize head
def __init__(self):
self.head = none
# recursive function to print
# linked list in reverse order
def printrev(self, temp):
if temp:
self.printrev(temp.next)
print(temp.data, end = ' ')
else:
return
# function to insert a new node
# at the beginning
def push(self, new_data):
new_node = node(new_data)
new_node.next = self.head
self.head = new_node
# driver code
llist = linkedlist()
llist.push(4)
llist.push(3)
llist.push(2)
llist.push(1)
llist.printrev(llist.head)
# this code is contributed by vinay kumar(vinaykumar71)
c
// c# program to print reverse of a linked list
using system;
public class linkedlist
{
node head; // head of list
/* linked list node*/
class node
{
public int data;
public node next;
public node(int d)
{
data = d; next = null;
}
}
/* function to print reverse of linked list */
void printreverse(node head)
{
if (head == null) return;
// print list of head node
printreverse(head.next);
// after everything else is printed
console.write(head.data " ");
}
/* utility functions */
/* inserts a new node at front of the list. */
public void push(int new_data)
{
/* 1 & 2: allocate the node &
put in the data*/
node new_node = new node(new_data);
/* 3\. make next of new node as head */
new_node.next = head;
/* 4\. move the head to point to new node */
head = new_node;
}
/*driver function to test the above methods*/
public static void main(string []args)
{
// let us create linked list 1->2->3->4
linkedlist llist = new linkedlist();
llist.push(4);
llist.push(3);
llist.push(2);
llist.push(1);
llist.printreverse(llist.head);
}
}
// this code has been contributed by rajput-ji
输出：
4 3 2 1
时间复杂度：o(n)
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