linked list c tutorial

typedef struct LINKEDLIST { int data; linkedlist* Pnext; }linkedlist; /** * Test 6, write a function to check whether there are loop in a linked list * @param plist [input] linked list * @return BOOL, if there is loop inside, return true, otherwise false */ BOOL Checkloop (linkedlist* plist) { linkedlist* P1 = plist;

->next = q;return head;}listnode* p = p1->next;listnode* q1 = q->next;Q->next = NULL;Doreverse (P);P1->next = q;P->next = Q1;return head;}}There are short implementations of code, but I am learning from others, execution speed is similar:listnode* dummy = new ListNode (0);Dummy->next = head;ListNode *lefti = dummy;for (;--m;--n) Lefti = lefti->next;ListNode *cur = lefti->next, *rtail = cur, *pre = nullptr;while (n--) {listnode* NXT = cur->next;Cur->next = pre;Pre = cur;cur = NXT;}Rtail->next = c

Title: Reverse list of nodes from m-n locationsFor example:Given1->2->3->4->5->null, m = 2 and n = 4,Return1->4->3->2->5->null.The nodes from the second to the fourth are reversed.where m and n satisfy the condition:1≤ m ≤ N ≤length of the list.Ideas:Still in reverse order, consider using a secondary space stack.The nodes of the m-n are sequentially placed into the stack and marked with the front and back two nodes adjacent to the stack node Pfirst an

Title DescriptionEnter a list that returns a ArrayList in the order of the list values from the end of the header.Title Addresshttps://www.nowcoder.com/practice/d0267f7f55b3412ba93bd35cfa8e8035?tpId=13tqId=11156tPage=1rp=2 ru=%2fta%2fcoding-interviewsqru=%2fta%2fcoding-interviews%2fquestion-rankingIdeasUsing the Python library function, create a new list, insert

I wrote an article a few days ago about how to create a linked list. Now, how can I print out content for another operation on the linked list.Keep up with the previous Code and write the print function.Void print (struct student * head){Struct student * p;Printf ("\ n these % d records are: \ n", n );P = head;If (head! = NULL)Do{Printf ("% ld % f \ n", p-> num,

","Zara","Yes","But","Row","OK","another"); linkednode"-------------------------------------------"); Dumplinkedlist (HEAD22);End Prepare System.out.println ("\n++++++++++++++++++++++++++++++++++++++++++\n");Start Test linkednode"-------------------------------------------"); linkednode/** recusive. Return head node as ascending. Would change parameters, non reentrant. */PublicStatic MergeSortedLinkedList1 (linkednodeif (A = =NULL) {return b; }if (b = =NULL) {return A; }if (Isfirstlessthansecond

Method 1: implement with the stack, step backward from the first node, and then push the node pointer (Address) into the stack in sequence. After all traversal ends, extract the content from the top of the stack and print the data. Since the later Node Address is above the stack, the printing order is printed from the end. Voidprint (node * List) {Stack Method 2: implement with recursionCodeConcise, but when the

Static linked list of adding and deleting, note the method of boundary check and parameter validity check. Test cases should be comprehensive!!!#include This article is from the "Small Stop" blog, please be sure to keep this source http://10541556.blog.51cto.com/10531556/1703847Linked list: static linked

"Idea": the direct addition of the test data is the line. But I still want to try the list, data structure after a year does write a little trouble. Alas.Note: The spelling of the node link. And p->p at the time of traversal, skimming the head node."AC Code":#include Blue Bridge Cup ALGO-107 9-7 linked list data summation operation (

Package Linklist.doublelinklist;public class Node {public int data;//data domain public node next;//node field, next node public node prve;// node field, previous node//head node Initialize public node (node next) {This.data=data;this.next=next;this.prve=null;} Non-head node initialization public node (int data,node next,node prve) {this.data=data;this.next=next;this.prve=prve;}}*************************************************************************************************************** ******

Single-cycle linked list and cyclic linked list 1 # include

#include C Language: "Single-linked list" To find the middle node of a single-linked list, requires only one time to traverse

New ListNode ( -1= dummy3; while (head2!=nullhead1!=null) { Pre.next=head1; = Head1.next; Head1.next=head2; = Head2.next; Head2.next=null; Pre=head2; Head1=head1next; Head2=head2next;}The code above is the correct notation.1ListNode dummy3 =NewListNode (-1);2ListNode pre =Dummy3;3 while(head2!=NULLhead1!=NULL){4pre.next=Head1;5Pre.next.next =head2;6 //ListNode head1next = head1.next;7 //head1.next=head2;8 //ListNode head2next = head2.next;9

Joseph's ring is an application of mathematics: n people (represented by numbers 1, 2, 3... n) are known to be sitting around a round table. The number of people numbered k starts to report, and the person counting to m is listed; his next person starts from 1 The number of people who count to m is displayed again. Repeat this rule until all the people around the Round Table are listed. Now we are writing a circular linked

The number of inserts is num. If it is smaller than the head, it is inserted in the chain table header. If it is larger than the elements in the chain table, it is inserted at the end; otherwise, it is inserted in the chain table; The Code is as follows: 1 link * link: add (link * plink, int num) {2 link * P1, * P2; 3 p1 = plink; 4 link * p0 = new link; 5 P0-> id = num; 6 7 while (P0-> ID> P1-> ID) (P1-> next! = NULL) {8 P2 = p1; 9 p1 = p1-> next; 10} 11 // insert 12 if (P0-> id 28 return pli

Classic question, the Code is as follows: 1 # include Given the head pointer of a linked list, it is required to traverse only once and reverse the element order in the single-linked list.

;Next; the } - } in if(NULL = =p1) theCur->next =P2; the if(NULL = =p2) AboutCur->next =P1; the the returnhead; the } + - //Recursive algorithm theNode *mergerecursive (node *head1, node *head2)Bayi { the if(NULL = =head1) the returnhead2; - if(NULL = =head2) - returnHead1; theNode *head =NULL; the the if(Head1->data data) the { -Head =Head1; theHead->next = Mergerecursive (head1->Next, head2); the } the Else94 { theHead =hea

Reverse a linked list from position m to n. Do it in-place and in one-pass. For example:Given1->2->3->4->5->NULL, M = 2 and n = 4, Return1->4->3->2->5->NULL. Note:Given m, n satisfy the following condition:1 ≤ m ≤ n ≤ length of list. It is still complicated to deal with this problem. Many boundary test cases need to be considered. My general idea is to mark the p

one, reverse single linked list Reverse single Chain table class ListNode {int val; ListNode next = null; ListNode (int val) {this.val = val; } public class Solution {public ListNode reverselist (ListNode head) {ListNode rehead=null; while (head!=null) {ListNode tmp=head; Head=head.next; Tmp.next=rehead; rehead=tmp; System.out.println (R

Solve the Joseph problem public class yuesefu{//define the nodes of the linked list public static class node{public int value; Node Next; public Node (int data) {this.value=data; }///Solve the Josephson problem/** head Ring List header node num reported number/public static node Yuesefu (node Head,int num) {if (head==null| | num