I understand that. I know it's just 2 nodes behind, but in terms of making the code go 2 nodes behind, that's a little tricky. Technically, the user needs to enter the length of the node and the size of the longlink - so it could be 20 nodes and longlink is every 6.
Here is my code to make it better to understand (Hopefully) Code: node *newnode = current; node *temp = current; node *start = current; int number = 0; int num = 0; for (int i = 1; i < length; i++) // this function is just ensuring the node before the deleted node points to 2 nodes in front { temp = temp -> shortlink; } if (newnode -> num == x) { number = current -> longlink -> num; // here as you can see i want to store the longlink value of the deleted node into the variable number current = current -> shortlink; temp -> shortlink = start -> shortlink; delete newnode; length--; } else while (newnode -> shortlink -> num != start -> num) { if (newnode -> shortlink -> num == x) { temp = newnode -> shortlink; number = temp -> longlink -> num; // here as you can see i want to store the longlink value of the deleted node into the variable number newnode -> shortlink = newnode -> shortlink -> shortlink; delete temp; length--; } newnode = newnode -> shortlink; } if (newnode -> shortlink -> shortlink -> num == length && newnode -> shortlink -> num == x) { delete newnode -> shortlink -> shortlink; newnode -> shortlink = start; length--; } for (int z = 1; z < length; z++) // this function goes through each node to see if any of the long links are pointing to the deleted node { // if i put a cout statement exactly here to cout number - it recognises the number variable fine if (longlinker -> longlink -> num == x) // if i've found it // once it looks into this if statement - number isn't recognised - why? { longlinker -> longlink -> num = number; // then i want to update the longlink of this which stores number } else { longlinker = longlinker -> shortlink; // otherwise keep iterating } }
I'm really stuck now some of the long links of the nodes that were referncing to the delete node now have ridiculous long link numbers (like 14155566)
The problem i got now is the following, and i'd REALLY be grateful if someone could assist me with this. Lets say i have 6 nodes and the long link is 3, then 1 points to 2 and 4, 2 points to 3 and 5, 3 points to 4 and 6 etc ..... When i delete node 1 and 2, then nodes 4 and 5 are longlinking back to 0 (for some reason). Can someone help me please? Code is here: Code: node *newnode = current; node *temp = current; node *start = current; node *longlinker = current; int number = 0; for (int i = 1; i < length; i++) { temp = temp -> shortlink; } if (newnode -> num == x) { start = start -> shortlink; number = current -> longlink -> num; current = current -> shortlink; temp -> shortlink = start; delete newnode; length--; cout << "longlink for 1 is" << number << endl; } else for (int p = 1 ; p < length; p++) { if (newnode -> shortlink -> num == x) { temp = newnode -> shortlink; number = temp -> longlink -> num; newnode -> shortlink = newnode -> shortlink -> shortlink; delete temp; length--; } else newnode = newnode -> shortlink; }
Still I dont get what problem you are facing. Its just the delete of a node from a link list then I have already given a code snippet to achieve the same and you can customize it to add extra information like short link and long link. Mine is just a simple where only one pointer exists and thats short link according to your problem and that is the next node.
Your code works fine (the sample), but that's only because each node is only linked to it's next node. My scenario is a little more complicated. Lets take this for instance, a list of 6 nodes and longlinks of 3: so the design looks like the following: 1 ----> 2( shortlink) && 1 ----> 4(longlink) 2 ----> 3( shortlink) && 1 ----> 5(longlink) 3 ----> 4( shortlink) && 1 ----> 6(longlink) 4 ----> 5( shortlink) && 1 ----> 1(longlink) 5 ----> 6( shortlink) && 1 ----> 2(longlink) 6 ----> 1( shortlink) && 1 ----> 3(longlink) Shortlinks are working fine. Now in that scenario, if i deleted node 1, that means nothing shall point to it . We must oberve though node 1 has a shortlink to 1 and a longlink to 4. Anyway, as you can see there, node 4 has a longlink to node 1, but node 1 is deleted, therefore this is wrong. So what i need to be able to do is, node 4 must point to the long link of the delete nodes long list. After deleting node 1, it should now look like 2 ----> 3( shortlink) && 1 ----> 5(longlink) 3 ----> 4( shortlink) && 1 ----> 6(longlink) 4 ----> 5( shortlink) && 1 ----> 4(longlink) ( 4's longlink must point to itself because 1's longlink was 4 5 ----> 6( shortlink) && 1 ----> 2(longlink) 6 ----> 2( shortlink) && 1 ----> 3(longlink) (notice 6's shortlink is now 2 - this works fine). My output is giving me this: 2 ----> 3( shortlink) && 1 ----> 5(longlink) 3 ----> 4( shortlink) && 1 ----> 6(longlink) 4 ----> 5( shortlink) && 1 ----> 0(longlink) (this should not be pointing to 0, it should be 4) 5 ----> 6( shortlink) && 1 ----> 2(longlink) 6 ----> 2( shortlink) && 1 ----> 3(longlink) Hopefully that clarifies my situation atm, and i'd really be grateful for some assistance.
Ok so it's all deleting the nodes fine and updating the links, except when i go to remove the 1st node - it has a problem. Could someone please examine what i'm doing wrong. Code: void ring::remove(int x) { node *newnode = current; node *temp = current; node *start = current; node *longlinker = NULL; node *shortdata = current; // this references to the first deleted node and will need it for testing longlink functionality int number = 0; for (int i = 1; i < length; i++) // this for loop is used for creating the short link - for .e.g if node 4 is deleted, then node 3 points to 5 and not 3 { temp = temp -> shortlink; } if (newnode -> num == x) { start = start -> shortlink; //number = current -> num; current = current -> shortlink; //number = temp -> num; temp = start; newnode = start; length--; } else for (int p = 1 ; p < length; p++) { if (newnode -> shortlink -> num == x) { temp = newnode -> shortlink; number = temp -> shortlink -> num; newnode -> shortlink = newnode -> shortlink -> shortlink; longlinker = newnode -> shortlink; //delete newnode; length--; } else newnode = newnode -> shortlink; } for (int y = 1; y <= length; y++) { if (longlinker -> longlink -> num == temp -> num) { longlinker -> longlink -> num = temp -> longlink -> num; } else longlinker = longlinker -> shortlink; } }
This sounds quite an interesting Linked list and so thought of programming it. Here is the code for you to see. Code: #include <stdio.h> #include <conio.h> #include <stdlib.h> #include <string.h> /* * Program for single linked list which inserts / deletes * data into linked list. The structure of linked list is * complex. Instead of just having a link to the shortlink * pointer it even stores a refernce to other link * * */ // Structure templates struct list { int no; struct list *shortlink; struct list *longlink; }; typedef struct list node; node *head = NULL; void init(node*); void InsertNode(); void DeleteNode(); void DisplayList(); void main() { char ch = 'n'; int opt; do { printf("\nEnter your option\n"); printf("\n1. Insert a node\n"); printf("\n2. Delete a particular node\n"); printf("\n3. Display all the nodes\n"); printf("\n4. Exit the application\n"); scanf("%d",&opt); switch(opt) { case 1: InsertNode(); DisplayList(); break; case 2: DeleteNode(); DisplayList(); break; case 3: DisplayList(); break; case 4: ch = 'y'; break; } }while(ch != 'y'); printf("\nDone by \"SHABBIR\"\n"); getch(); } void init(node *current) { printf("\nEnter number\n"); scanf("%d",¤t->no); current->shortlink = NULL; current->longlink = NULL; } void InsertNode() { node *newnode; newnode=(node*)malloc(sizeof(node)); init(newnode); if(head == NULL) { head = newnode; } else { node *current = head; node *longlinkedNode = NULL; // Move to the end of the link list for(int i = 1; current->shortlink!=NULL ; i++) { if(i % 2 == 0) { longlinkedNode = current; } current=current->shortlink; } if(current->shortlink==NULL) { newnode->shortlink = current->shortlink; newnode->longlink = longlinkedNode; current->shortlink = newnode; } } } void DeleteNode() { node* current = head; int rno; node *newnode,*temp; printf("\nEnter the number whose node you want to delete\n"); scanf("%d",&rno); newnode=current; if(current->no==rno) { // Deleting the first node newnode=current; current=current->shortlink; free(newnode); head = current; return; } else { node* longlinkedNode = NULL; for(int i = 1; newnode->shortlink->shortlink!=NULL; i++) { if(i % 2 == 0) { longlinkedNode = current; } // Checking condition for deletion of // all nodes except first and last node if(newnode->shortlink->no==rno) { temp=newnode->shortlink; newnode->shortlink=newnode->shortlink->shortlink; longlinkedNode->longlink = temp->longlink; free(temp); head = current; return; } newnode=newnode->shortlink; } if(newnode->shortlink->shortlink==NULL && newnode->shortlink->no==rno) { // Checking condition for deletion of last node free(newnode->shortlink); longlinkedNode->longlink = newnode->shortlink->longlink; newnode->shortlink=NULL; head = current; return; } } } void DisplayList() { node* current = head; while(current!=NULL) { printf("\n %d ",current->no); current=current->shortlink; } } I dont think it was that different a code from what I have posted.
Hell o I am new member of this site. I am having some problems with linked lists. I want the solution for the problem described below can you help me with the code . We have to open a new text file and do these operations in that for eg:Name and mark of a student Add Node . Delete Node - Insert Node - Same as Add Node, but takes an additional parameter of the location in which this new node is to be inserted Sort Nodes - Display Nodes - Should display the whole list in Name1 (mark), Name 2 (mark)... format. Save Nodes - This will output the curent list and save it into the same mark.txt file Exit - Exits the Application. I need it on Wednesday
Hi, This post is very informative, however I would like some specific information. If someone can help me then please send me a private message. Best Regards,
