[Programming question] enter a one-way linked list and output the last K nodes in the list.

13th questions (linked list ):
Question: enter a one-way linked list and output the k-th node in the list. The last 0th nodes of the linked list are linked lists.
.
The linked list node is defined as follows:
Struct listnode
{
Int m_nkey;
Listnode * m_pnext;
};

 

My idea: first flip the linked list, and then from the flip of the linked list head to the end of the number of K-1, return, flip the linked list again.

The Code is as follows: note that this idea is very poor. The reason for the difference is: If you only use the original method, first traverse the count, and then traverse and find the last K, you need to traverse twice. However, in my mind, if I flip the linked list twice, I have to traverse it twice. Also go to the K-1 step to find the last K. Tedious.

/* 13th question (linked list): Question: enter a one-way linked list and output the k-th node in the list. The last 0th nodes of the linked list are the tail pointer of the linked list. The linked list node is defined as follows: struct listnode {int m_nkey; listnode * m_pnext;}; Start time = 20: 40end time */# include <stdio. h> # include <stdlib. h> typedef struct listnode {int m_nkey; listnode * m_pnext;} listnode; listnode * reverselist (listnode * l) {listnode * PP = L; listnode * P = L-> m_pnext; PP-> m_pnext = NULL; while (P! = NULL) {listnode * P3 = p-> m_pnext; P-> m_pnext = pp; pp = P; P = P3;} return pp ;} listnode * getkthnodefromback (listnode * l, int K) {If (k = 0) return NULL; listnode * RL = reverselist (l); listnode * P = RL; for (INT I = 0; P! = NULL & I <K-1; I ++) // note that the detection K cannot exceed the length of the linked list, otherwise null {P = p-> m_pnext;} reverselist (RL) is returned ); return P;} void displaylist (listnode * l) {listnode * P = L; while (P! = NULL) {printf ("% d", p-> m_nkey); P = p-> m_pnext;} printf ("\ n ");} void createlist (listnode * & L) {int D; scanf ("% d", & D); If (D! = 0) {L = (listnode *) malloc (sizeof (listnode); L-> m_nkey = D; L-> m_pnext = NULL; createlist (L-> m_pnext) ;}} int main () {listnode * l = NULL; createlist (l); listnode * PK = getkthnodefromback (L, 3 ); displaylist (l); Return 0 ;}

 

Online thinking: http://blog.sina.com.cn/s/blog_60c8379d01013z0s.html

Analysis: Use two pointers: low and fast. First, point the fast pointer to the K element, and then low and fast traverse backward at the same time. When fast traverses to the end, low traverses the K at the bottom. Check whether the boundary and pointer are empty. Check whether K exceeds the length of the linked list.

LinkList.h #ifndef LINKLIST_H#define LINKLIST_H#include <stdlib.h>#include <stdio.h>#include <iostream>using namespace std;typedef struct node{int val;struct node *next;}node,*pNode;pNode CreateLinkList(pNode &T);pNode FindLastKth(pNode T,int k);#endifLinkList.cpp#include "LinkList.h"pNode CreateLinkList(pNode &T){int ival;cin>>ival;if(ival==0)return NULL;T = (pNode)malloc(sizeof(node));T->val = ival;T->next = NULL;T->next = CreateLinkList(T->next);return T;}pNode FindLastKth(pNode T,int k){pNode low=T,fast = T;for(int i=0;i<k&&fast!=NULL;i++){fast = fast->next;}if(fast == NULL)return NULL;while(fast!=NULL){fast = fast->next;low = low->next;}return low;}
Main.cpp#include "LinkList.h"void main(){pNode T;T = NULL;CreateLinkList(T);int k=3;pNode pnode;pnode = FindLastKth(T,k);cout<<pnode->val;system("pause");}

 

[Programming question] enter a one-way linked list and output the last K nodes in the list.