best pen for penultimate

Title: Enter a linked list to output the list of the lowest K-points in the list.In order to conform to most people's habit, the subject starts from 1, that is, the tail node of the list is the 1th node in the bottom.For example, a list of 6 nodes, starting from the beginning, their values are 1,2,3,4,5,6. The bottom 3rd node of the list is a node with a value of 4.In order to get the K-node, it is natural to go to the end of the list and back to K-step from the tail. But we see from the definit

1. Topics Given a linked list, deletes the penultimate node of the linked list and returns the head node of the linked list. Example:Given a linked list: 1->2->3->4->5, and n = 2.When the second-to-last node is deleted, the list becomes 1->2->3->5. DescriptionThe given n guarantee is valid. Advanced:Can you try to use a scan to implement it? 2. Ideas Defines two pointers P1, p2, and j

"Huawei OJ" "Algorithm Total chapter" "Huawei OJ" "035-Output unidirectional link list of the penultimate K nodes" "Project download" title description输入一个单向链表，输出该链表中倒数第k个结点，链表的倒数第0个结点为链表的尾指针。Enter a description输入说明1 输入链表结点个数2 输入链表的值3 输入k的值Output description输出一个整数Input example8 1 2 3 4 5 6 7 8 4Output example4Algorithm implementationImportOrg.omg.SendingContext.RunTime;ImportJava.util.List;ImportJava.util.Scanner;/** * Author: Wang Junshu * date:2015-

Title: Enter the single-linked table L of the lead node and output the reciprocal K-node in the single-linked list. The last No. 0 node of a single linked list is the tail pointer of the single linked list. Requires only one single linked list to traverse once.Problem Solving Ideas:If it is not required to traverse only one single linked list, we can first go through a single linked list, to find its total number of nodes N (including the head node), so the node of the single-linked list is from

Idea 1: Two pointers P1, p2, p1 traverse from the beginning, when the P1 reaches the K-node, p2 begins; When P1 reaches the last node in the list, p2 refers to the node in the list that is the reciprocal K-node. Public classnode{ Public intdata; PublicNode Next; PublicNode (intdata) { This. data =data; } PublicNode Removelastknode (node head,intk) { if(Head = =NULL|| K ){ returnHead; } Node p1=Head; Node P2=Head; intnum = 1; while(P1.next! =NULL Num k) {

); + Break ; - } the i++;Bayi head = head->next; the } the } - } - */ the //idea two: the voidLooknode (NODE *head,intLenintk) the { the inti =K; -NODE * p1,*P2; theP1 = P2 =head; the if(k>len| | k0) the {94printf"The linked list is empty, or the node for the query does not exist. There is no reciprocal%d node \ n", k); the return ; the } the 98 //P2 points to the first K-node About while(k>=1) - {101P2 = p2->Next;102k--;103 }1

Problem Description:Find the countdown k node in the listThinking Analysis:With two pointers, one after the other, keep the k distance, the front pointer moves to the end, the back pointer is just right up to the K-node,Consider the case where K is 0 and the penultimate K-node does not exist.Reference code:listnode* findkthtotail (ListNode * phead,unsigned int k){if (NULL = = Phead | | k = = 0){return NULL;}ListNode *pahead = Phead;for (unsigned int i

Enter a one-way list to output the last K nodes in the linked list. The bottom No. 0 node of the list is the tail pointer of the linked list. My idea is 2 pointers to the back, you think Ah, if it is a pointer, it must traverse 2 times, the first traversal total number of nodes, the second time to traverse the final result//Such a practice is obviously not good enough, time complexity into 2n, but if we use 2 pointers, the distance between them K nodes, When one node arrives at null//(tail), the

When we do the algorithm, we are more or less confronted with the problem that we need to get the number of the reciprocal or positive numbers of a data set. So let's take a look at this question today and how to get to the bottom K node. When we get this problem, we naturally think that we let the list start at the end of the next K-1 times is not the first K-1 node, but must note that this is a one-way list. Then the solution to the idea may be divided, most people will think of us to traverse

Title: Enter a linked list. Output the list of the countdown K-junction.In order to conform to the habits of most people, the subject starts counting from 1. That is, the tail node of the list is the bottom 1th node.For example, a list of 6 nodes. Starting from the beginning, their values are sequentially 1. 2. 3,4,5,6. The 3rd node in the list is a node with a value of 4.To get the K-node, it is very natural to go first to the end of the list. Then go back from the tail to the K-step.But we see

Title Description:Enter a list to output the last K nodes in the linked list.Ideas:First check the legitimacy of the parameters, Head==null or the number of nodes less than K directly return NULL.Let head advance k-1 step, ans points to the head node, then head forward one step, ans also step forward. When the head reaches the last node, ans points to the reciprocal K-node. Time complexity O (n).Solve:1 /*2 Public class ListNode {3 int val;4 ListNode next = null;5 6 listnode (int val) {7 this.va

pointer moves forward k-1. for(inti=0; i) { //if K is greater than the number of nodes, next May point to NULL, return null if(Pforward.next = =NULL) return NULL; Pforward=Pforward.next; } //The second pointer does not move temporarilyPbehind =Head; //Two pointers move together//when the previous pointer traverses to the last node, a pointer to the bottom of the K-node while(Pforward.next! =NULL) {Pforward=Pforward.next; Pbehind=Pbehin

Note the typical example of robustness and algorithmic efficiency: (header file omitted)typedefstructnode{intdata; structnode*Next;} ListNode; ListNode* Findkthtotail (listnode* plisthead, unsignedintk);intMain () {inti; Const intN = -; ListNode*head, *p; //random number seedSrand ((unsignedint) Time (0)); //Generate linked listHead = P =NewListNode; Head->data = rand ()%101; Head->next =nullptr; for(i =0; i ) {p= P->next =NewListNode; P->data = rand ()%101; P->next =nullptr; } //Output Link

1, linked list definitiontypedef struct LISTELEMENT_T_ { void *data; struct Listelement_t_ *next;} listelement_t;typedef struct list_t_{ int size; int capacity; listelement_t *head; listelement_t *tail;} list_t;2. Find the K node data of the linked listSet the speed pointer, the fast pointer ahead of the slow pointer K-1 nodes, and then the speed pointer and then traverse the linked list, when the fast pointer over the end of the list, the slow pointer is the bottom of the K no

) {System.out.print (P.m_nvalue+" "); P=P.m_pnext; } System.out.println ("\ n"); } //the reciprocal k nodes of the output and output list Public StaticListNode Printktotail (ListNode phead,intk) {if(phead==NULL|| K) return NULL; ListNode Pahead=Phead; ListNode Ptail=NULL; for(inti=0;i) { if(pahead.m_pnext!=NULL) Pahead=Pahead.m_pnext; Else return NULL; } ptail=Phead; while(pahead.m_pnext!=NULL) {Pahead=Pahead.m_pnext; Ptail=Ptail.m_pnext; }

actually the same. 1.1.1 Find the Countdown k node Idea one Go through the list first, get the link table length, find the position of the penultimate K, and then start traversing from the list head, so we have a total of (2l-k) nodes. Idea two And find the middle point of thinking a little bit the same, just let a pointer go first k nodes, and then two pointers each time a node, when the front of the pointer to reach the end point, the back of

Title: Enter a list, output the list of the countdown K nodes, in order to conform to the majority of the habit, the topic from 1 to count, that is, the tail node of the list is the penultimate node.Method 1: The list is traversed first to get the number of the list n, the penultimate K node is n-k+1 and then traverse the list, find the N-k+1 node is the penultimate