linked list c tutorial

Topic:Given A linked list, determine if it has a cycle in it.Ideas:For judging whether the chain list has a ring, the method is very simple, with two pointers, beginning to point to the head node, one is a fast pointer, two steps at a time, a slow pointer, one step at a time, when two pointers coincide with the existence of the ring.Fast first into the ring, afte

Flatten a binary tree to a fake "linked list" in pre-order traversal.Here we use the right pointer in TreeNode as the next pointer in ListNode.NoticeDon ' t forget to mark the left child of each node to null. Or you'll get time limit exceeded or Memory Limit exceeded.Example 1 1 2 / \ 2 5 => 3 / \ \ 3 4 6 4

This problem requires a comparison of the pointers to understand the position. Then the method is straightforward: the odd number of connected, even the even, and finally put together./** Definition for singly-linked list. * struct ListNode {* int val; * ListNode *next; * ListNode (int x) : Val (x), Next (NULL) {}}; */classSolution { Public: ListNode* Oddevenlist (listnode*head) { if(head==null| | he

translation给定一个访问节点的路径，写一个函数去删除在一个单向链表中除尾部以外的节点。假设这个链表是1234，并且你被给予了第3个值为3的节点，那么在调用你的函数之后这个链表应该变为124。OriginalWrite a function toDelete a node (except theTailinchA singly linkedList,givenOnly access to thatNode. Supposed theLinkedList is 1-2-3-4 andYou aregiven the ThirdNode withValue3, theLinkedListShould become1-2-4 AfterCalling your function.Code/*** Definition for singly-linked list.* struct ListNode

Title meaning: If there is a ring, return to the entry node.Idea: First judge have no ring, and then calculate the knot number of the ring, then P1 point to the head, P2 back to the node, p1, p2 meet for the entrance nodePS: or the idea of using the pointer spacing1 /**2 * Definition for singly-linked list.3 * struct ListNode {4 * int val;5 * ListNode *next;6 * ListNode (int x): Val (x), Next (NULL) {}7 *

Code:/** * Definition for singly-linked list. * public class ListNode {* int val; * ListNode Next; * listnode (int x) {val = x;}}} */public class Soluti On {public void Deletenode (ListNode node) { if (node = = null) return; while (node.next! = null) { node.val = node.next.val; if (node.next.next! = null) { node = node.next; }

Package Linklist;public class NodeGorgeous Split-line ********************************************************* ***************Package linklist;/** * **************** single-linked list with lead node and its implementation ********************* * * @author WL * */public class SINGLYLINKLISTL T t> {nodeJava data structure System--linked

A long time ago practice, it is not very difficult to see. Packageproject;classnode{Private intId//Private is only accessible to this class of objects and methods. 　　 PrivateString name; PublicNode Next;//point to next class node 　　 PublicNode (intId,string name)//a method for constructing a parameter{ This. id=ID; This. name=name; } Public voidDisplayLink ()//Display node content{System.out.println ("ID:" +id+ "" + "Name:" +name); }}classlinklist{PrivateNode first; Publiclinklist () { F

The Code is a non-cyclic two-way linked list, including node insertion and deletion operations [cpp] // LinkList_D.cpp: Defines the entry point for the console application. // # include "stdafx. h "# include" stdio. h "# include" stdlib. h "# include" iostream "using namespace std; typedef int DataType; typedef struct DLNode {DataType data; struct DLNode * prior; struct DLNode * next;} DLNode, * DLinkList;

Topic:Reverse a linked list from position m to N. Do it in-place and in One-pass.For example:Given 1->2->3->4->5->NULL , m = 2 and n = 4,Return 1->4->3->2->5->NULL .Note:Given m, n satisfy the following condition:1 ≤ m ≤ n ≤length of list.Ideas:Find the M node, reverse the pointer from node m to N, and then concatenate the flip-up/** Definition for singly-linked

Iteration:Code:/*** Definition for singly-linked list.* Public class ListNode {* int val;* ListNode Next;* ListNode (int x) {val = x;}* }*/public class Solution {Public ListNode reverselist (ListNode head) {if (head = = NULL) return null;ListNode tail = head.next;ListNode cur = head;while (tail!= null) {ListNode NEX = Tail.next;Tail.next = head;Cur.next = NEX;head = tail;tail = NEX;}return head;}}Recursion:

1. Node The storage unit used to store a single data in the linked list. A linked list must be composed of at least two parts: the data domain and the pointer domain. Generally, nodes are defined: Struct Node { ELEM data; // The ELEM type refers to the basic data type. Struct node * next; } Typedef struct node elemsn;

/** Definition for singly-linked list. * struct ListNode {* int val; * struct ListNode *next; *};*/structlistnode* Reverselist (structlistnode*head) { if(!head| |! Head->next)returnHead; structListNode *cur=Head; structListNode *tail=Head; while(tail->next) Tail=tail->Next; while(cur!=tail) {Head=cur->Next; Cur->next=tail->Next; Tail->next=cur; Cur=Head; } returnhead;}There is a problem, leetcode gi

; } Else if(POS >= m Pos N) {Pnext= P-Next; PNext =PLast; if(pos = =m) {PLastNext =NULL; NEWM=PLast; } pLast=p; if(pos = = N-1) {newn=p; Aftern=Pnext; } P=Pnext; }Else{pLast=p; P= P-Next; } POS++; } if(m==1 Aftern = =NULL) {Head=newn; }Else if(M = =1) {Head=newn; NEWMNext =Aftern; }Else{BeforemNext =newn; NEWMNext =Aftern; } returnHead; } ListNode* Reverselist (listnode*head) { if(head = = NULL)returnHead; ListNode*plast = head, *p = Head, Next, *pnext =NULL; wh

Package Linklist.onewaycircular;public class Node {public int data;public node next;//header node initialize public node (node next) { This.next=next;} General node Initialization public node (int data,node next) {This.data=data;this.next=next;}}I'm a gorgeous cut *************************************************************************. ******Package Linklist.onewaycircular;public class Onewaycircularlinklist {node head;//header node current;//current node int size;//

Given a linked list, return the node where the cycle begins. If There is no cycle, return null .Follow up:Can you solve it without using extra space?The first question asked whether there is a ring can be seen using a fast pointer as long as there is a ring will meet the other from the meeting point can be seen from the point of the distance between the distance ring and head to the starting point of the d

1#include"head.h"2 structStudent *creat ()3 {4 structStudent *head, *P1, *P2;//first open three struct pointers, *head, (as the return head pointer)5P1 = P2 = (structStudent *)malloc(LEN);6scanf_s ("%s%f", P1->num, N, p1->score);//first read the input information, according to read the information to Judge7head = NULL;//make the head pointer point to the null pointer first8n =0;//to count the number of members in a list9 while(strcmp (P1->num,"0") !=0)//according to the information read

1#include"head.h"2 structStudent *del_same_age (structStudent*head,intAge )3 {4 structStudent *p, *pt;intFind =0;5p = pt =head;6 while(P! =NULL)//when looping to the last node7 {8 9 if(P->stu_age = =Age )//if equalTen { Onefind++;//description found A if(p = =Head )//If found to be the first node - { -Head = p->Next; thep = p->next;//loop node move back -PT =p; - } - Else//If not the head node + { -Pt->next

. prev =NULL; }; varLength = 0; varHead =NULL; varTail =NULL; This. Insert =function(position, Element) {if(Position >=0 position length) { varnode =NewNode (Element), current=Head, Previous, index= 0; if(Position = = 0){ if(!head) {Head=node; Tail=node; } Else{Node.next=Current ; Current.prev=node; Head=node; } } Else if(Position = = =length) { Current=tail; Current.next=node; Node.prev=Current ; Tail=node; } Else { while(Index++ positio

#include C language: "Single linked list" reverse inverted single linked list