一个有趣的打印链表反向的方法
我们得到一个链表，我们需要以相反的顺序打印链表。
例子：
Input : list : 5-> 15-> 20-> 25
Output : Reversed Linked list : 25-> 20-> 15-> 5
Input : list : 85-> 15-> 4-> 20
Output : Reversed Linked list : 20-> 4-> 15-> 85
Input : list : 85
Output : Reversed Linked list : 85为了以相反的顺序打印列表，我们已经讨论了反向迭代和递归方法。
在这篇文章中，讨论了一个有趣的方法，它不需要递归，也不需要修改列表。该函数也只访问链表的每个节点一次。
技巧：在没有任何递归函数或循环的情况下以相反的顺序打印列表背后的想法是使用回车符（“r”）。为此，我们应该了解列表的长度。现在，我们应该打印 n-1 个空格，然后打印第一个元素，然后是“r”，再打印 n-2 个空格和第二个节点，然后是“r”，依此类推。
回车（“r”）：它命令打印机（光标或系统控制台的显示器）将光标的位置移动到同一行的第一个位置。
C/C++
// C program to print reverse of list
#include
#include
/* Link list node */
struct Node {
int data;
struct Node* next;
};
/* Function to reverse the linked list */
void printReverse(struct Node** head_ref, int n)
{
int j = 0;
struct Node* current = *head_ref;
while (current != NULL) {
// For each node, print proper number
// of spaces before printing it
for (int i = 0; i < 2 * (n - j); i++)
printf(" ");
// use of carriage return to move back
// and print.
printf("%d\r", current->data);
current = current->next;
j++;
}
}
/* Function to push a node */
void push(struct Node** head_ref, int new_data)
{
struct Node* new_node =
(struct Node*)malloc(sizeof(struct Node));
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
/* Function to print linked list and find its
length */
int printList(struct Node* head)
{
// i for finding length of list
int i = 0;
struct Node* temp = head;
while (temp != NULL) {
printf("%d ", temp->data);
temp = temp->next;
i++;
}
return i;
}
/* Driver program to test above function*/
int main()
{
/* Start with the empty list */
struct Node* head = NULL;
// list nodes are as 6 5 4 3 2 1
push(&head, 1);
push(&head, 2);
push(&head, 3);
push(&head, 4);
push(&head, 5);
push(&head, 6);
printf("Given linked list:\n");
// printlist print the list and
// return the size of list
int n = printList(head);
// print reverse list with help
// of carriage return function
printf("\nReversed Linked list:\n");
printReverse(&head, n);
printf("\n");
return 0;
}
Java
// Java program to print reverse of list
import java.io.*;
import java.util.*;
// Represents node of a linkedlist
class Node {
int data;
Node next;
Node(int val)
{
data = val;
next = null;
}
}
public class GFG
{
/* Function to reverse the linked list */
static void printReverse(Node head, int n)
{
int j = 0;
Node current = head;
while (current != null) {
// For each node, print proper number
// of spaces before printing it
for (int i = 0; i < 2 * (n - j); i++)
System.out.print(" ");
// use of carriage return to move back
// and print.
System.out.print("\r" + current.data);
current = current.next;
j++;
}
}
/* Function to push a node */
static Node push(Node head, int data)
{
Node new_node = new Node(data);
new_node.next = head;
head = new_node;
return head;
}
/* Function to print linked list and find its
length */
static int printList(Node head)
{
// i for finding length of list
int i = 0;
Node temp = head;
while (temp != null)
{
System.out.print(temp.data + " ");
temp = temp.next;
i++;
}
return i;
}
// Driver code
public static void main(String args[])
{
/* Start with the empty list */
Node head = null;
// list nodes are as 6 5 4 3 2 1
head = push(head, 1);
head = push(head, 2);
head = push(head, 3);
head = push(head, 4);
head = push(head, 5);
head = push(head, 6);
System.out.println("Given linked list: ");
// printlist print the list and
// return the size of list
int n = printList(head);
// print reverse list with help
// of carriage return function
System.out.println("Reversed Linked list: ");
printReverse(head, n);
System.out.println();
}
}
// This code is contributed by rachana somaC#
// C# program to print reverse of list
using System;
// Represents node of a linkedlist
public class Node {
public int data;
public Node next;
public Node(int val)
{
data = val;
next = null;
}
}
public class GFG
{
/* Function to reverse the linked list */
static void printReverse(Node head, int n)
{
int j = 0;
Node current = head;
while (current != null) {
// For each node, print proper number
// of spaces before printing it
for (int i = 0; i < 2 * (n - j); i++)
Console.Write(" ");
// use of carriage return to move back
// and print.
Console.Write("\r" + current.data);
current = current.next;
j++;
}
}
/* Function to push a node */
static Node push(Node head, int data)
{
Node new_node = new Node(data);
new_node.next = head;
head = new_node;
return head;
}
/* Function to print linked list and find its
length */
static int printList(Node head)
{
// i for finding length of list
int i = 0;
Node temp = head;
while (temp != null)
{
Console.Write(temp.data + " ");
temp = temp.next;
i++;
}
return i;
}
// Driver code
public static void Main(String []args)
{
/* Start with the empty list */
Node head = null;
// list nodes are as 6 5 4 3 2 1
head = push(head, 1);
head = push(head, 2);
head = push(head, 3);
head = push(head, 4);
head = push(head, 5);
head = push(head, 6);
Console.WriteLine("Given linked list: ");
// printlist print the list and
// return the size of list
int n = printList(head);
// print reverse list with help
// of carriage return function
Console.WriteLine("Reversed Linked list: ");
printReverse(head, n);
Console.WriteLine();
}
}
// This code is contributed by Arnab KunduPython3
# Python3 program to print reverse of list
# Link list node
class Node:
def __init__(self):
self.data= 0
self.next=None
# Function to reverse the linked list
def printReverse( head_ref, n):
j = 0
current = head_ref
while (current != None):
i = 0
# For each node, print proper number
# of spaces before printing it
while ( i < 2 * (n - j) ):
print(end=" ")
i = i + 1
# use of carriage return to move back
# and print.
print( current.data, end = "\r")
current = current.next
j = j + 1
# Function to push a node
def push( head_ref, new_data):
new_node = Node()
new_node.data = new_data
new_node.next = (head_ref)
(head_ref) = new_node
return head_ref;
# Function to print linked list and find its
# length
def printList( head):
# i for finding length of list
i = 0
temp = head
while (temp != None):
print( temp.data,end = " ")
temp = temp.next
i = i + 1
return i
# Driver program to test above function
# Start with the empty list
head = None
# list nodes are as 6 5 4 3 2 1
head = push(head, 1)
head = push(head, 2)
head = push(head, 3)
head = push(head, 4)
head = push(head, 5)
head = push(head, 6)
print("Given linked list:")
# printlist print the list and
# return the size of list
n = printList(head)
# print reverse list with help
# of carriage return function
print("\nReversed Linked list:")
printReverse(head, n)
print()
# This code is contributed by Arnab Kundu输出：
Given linked list:
6 5 4 3 2 1
Reversed Linked List:
1 2 3 4 5 6输入和输出图示：
输入：6 5 4 3 2 1
第一次迭代 _ _ _ _ _ 6
第二次迭代 _ _ _ _ 5 6
第三次迭代 _ _ _ 4 5 6
第四次迭代 _ _ 3 4 5 6
第 5 次迭代 _ 2 3 4 5 6
最终输出 1 2 3 4 5 6
注意：上述程序可能无法在在线编译器上运行，因为它们不支持控制台上的回车等任何内容。
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