原文：
给定一个包含n个节点和正整数k的链表应小于或等于n。任务是打印该节点的最后k个节点。 以相反的顺序列出。
示例：
input: list: 1->2->3->4->5, k = 2
output: 5 4
input: list: 3->10->6->9->12->2->8, k = 4
output: 8 2 12 9
来源：。
递归方法：递归遍历链表。 从每个递归调用返回时，请跟踪节点编号，将最后一个节点视为编号 1，将倒数第二个节点视为编号 2，依此类推。 可以借助全局变量或指针变量来跟踪此计数。 借助此count变量，打印节点号小于或等于k的节点。
下面是上述方法的实现：
c
// c implementation to print the last k nodes
// of linked list in reverse order
#include
using namespace std;
// structure of a node
struct node {
int data;
node* next;
};
// function to get a new node
node* getnode(int data)
{
// allocate space
node* newnode = new node;
// put in data
newnode->data = data;
newnode->next = null;
return newnode;
}
// function to print the last k nodes
// of linked list in reverse order
void printlastkrev(node* head,
int& count, int k)
{
// if list is empty
if (!head)
return;
// recursive call with the next node
// of the list
printlastkrev(head->next, count, k);
// count variable to keep track of
// the last k nodes
count ;
// print data
if (count <= k)
cout << head->data << " ";
}
// driver code
int main()
{
// create list: 1->2->3->4->5
node* head = getnode(1);
head->next = getnode(2);
head->next->next = getnode(3);
head->next->next->next = getnode(4);
head->next->next->next->next = getnode(5);
int k = 4, count = 0;
// print the last k nodes
printlastkrev(head, count, k);
return 0;
}
java
// java implementation to print the last k nodes
// of linked list in reverse order
class gfg
{
// structure of a node
static class node
{
int data;
node next;
}
// function to get a new node
static node getnode(int data)
{
// allocate space
node newnode = new node();
// put in data
newnode.data = data;
newnode.next = null;
return newnode;
}
static class c
{
int count = 0;
}
// function to print the last k nodes
// of linked list in reverse order
static void printlastkrev(node head, c c, int k)
{
// if list is empty
if (head == null)
return;
// recursive call with the next node
// of the list
printlastkrev(head.next, c, k);
// count variable to keep track of
// the last k nodes
c.count ;
// print data
if (c.count <= k)
system.out.print(head.data " ");
}
// driver code
public static void main(string[] args)
{
// create list: 1->2->3->4->5
node head = getnode(1);
head.next = getnode(2);
head.next.next = getnode(3);
head.next.next.next = getnode(4);
head.next.next.next.next = getnode(5);
int k = 4;
c c = new c();
// print the last k nodes
printlastkrev(head, c, k);
}
}
// this code is contributed by prerna saini
python
# python implementation to print the last k nodes
# of linked list in reverse order
# node class
class node:
# function to initialise the node object
def __init__(self, data):
self.data = data # assign data
self.next =none
# function to get a new node
def getnode(data):
# allocate space
newnode = node(0)
# put in data
newnode.data = data
newnode.next = none
return newnode
class c:
def __init__(self, data):
self.count = data
# function to print the last k nodes
# of linked list in reverse order
def printlastkrev(head, c, k):
# if list is empty
if (head == none):
return
# recursive call with the next node
# of the list
printlastkrev(head.next, c, k)
# count variable to keep track of
# the last k nodes
c.count = c.count 1
# print data
if (c.count <= k) :
print(head.data, end = " ")
# driver code
# create list: 1->2->3->4->5
head = getnode(1)
head.next = getnode(2)
head.next.next = getnode(3)
head.next.next.next = getnode(4)
head.next.next.next.next = getnode(5)
k = 4
c = c(0)
# print the last k nodes
printlastkrev(head, c, k)
# this code is contributed by arnab kundu
c
// c# implementation to print the last k
// nodes of linked list in reverse order
using system;
class gfg
{
// structure of a node
public class node
{
public int data;
public node next;
}
// function to get a new node
static node getnode(int data)
{
// allocate space
node newnode = new node();
// put in data
newnode.data = data;
newnode.next = null;
return newnode;
}
public class c
{
public int count = 0;
}
// function to print the last k nodes
// of linked list in reverse order
static void printlastkrev(node head, c c, int k)
{
// if list is empty
if (head == null)
return;
// recursive call with the next
// node of the list
printlastkrev(head.next, c, k);
// count variable to keep track
// of the last k nodes
c.count ;
// print data
if (c.count <= k)
console.write(head.data " ");
}
// driver code
public static void main(string []args)
{
// create list: 1->2->3->4->5
node head = getnode(1);
head.next = getnode(2);
head.next.next = getnode(3);
head.next.next.next = getnode(4);
head.next.next.next.next = getnode(5);
int k = 4;
c c = new c();
// print the last k nodes
printlastkrev(head, c, k);
}
}
// this code is contributed by arnab kundu
输出：
5 4 3 2
时间复杂度：o(n)。
迭代方法：想法是使用。
-
将所有链表节点推入栈。
-
从栈中弹出
k个节点并进行打印。
时间复杂度：o(n)。
双指针方法的想法类似于从的末尾找到第k个节点。
-
将第一个指针向前移动
k个节点。 -
现在从头开始第二个指针。
-
当第一个指针到达末尾时，第二个指针指向第
k个节点。 -
最后使用第二个指针，打印最后
k个节点。
时间复杂度：o(n)。
如果您喜欢 geeksforgeeks 并希望做出贡献，则还可以使用 tribution.geeksforgeeks.org 撰写文章，或将您的文章邮寄至 tribution@geeksforgeeks.org。 查看您的文章出现在 geeksforgeeks pg电子试玩链接主页上，并帮助其他 geeks。
如果您发现任何不正确的地方，请单击下面的“改进文章”按钮，以改进本文。
麻将胡了pg电子网站的版权属于：月萌api www.moonapi.com，转载请注明出处
