singly-linked-list

Here is my struct struct ListItem{ int data; struct ListItem *next; }; Assuming the first node of the linked list will have data = 0, I want to write a for loop that creates a linked list of size 5 but I'm not sure how to work I tried the following int main(int argc, char* argv[]){ struct ListItem a; a.data = 0; for (int i = 1; i < 5; i++){ struct ListItem *pointer = &a; struct ListItem nextnode; nextnode.data = i; a.next = &nextnode; pointer = pointer->next; } } But the result is a.data = 0 and a.next->data = 4 Don't modify a. Take a temp node starting as a. Make it's next point to the new

TL; DR After reading the passage about persistence in Okasaki's Purely Functional Data Structures and going over his illustrative examples about singly linked lists (which is how Haskell's lists are implemented), I was left wondering about the space complexities of Data.List 's inits and tails ... It seems to me that the space complexity of tails is linear in the length of its argument, and the space complexity of inits is quadratic in the length of its argument, but a simple benchmark indicates otherwise. Rationale With tails , the original list can be shared. Computing tails xs simply

While trying to implement a simple singly linked list in C#, I noticed that == does not work while comparing two object type variables boxed with an int value but .Equals works. Wanted to check why that is so. The below snippet is a generic object type Data property public class Node { /// <summary> /// Data contained in the node /// </summary> private object Data { get; set; }; } The below code traverses the singly linked list and searches for a value of type object - /// <summary> /// <param name="d">Data to be searched in all the nodes of a singly linked list /// Traverses through each node

I am just curious about node . node.next and node.next. Until now we have read that node has two part in which it store data and address of next node(node.next) then whats about node.next.next. ? Where the address of node.next.next will be stored. I just implemented node.next.next to delete a number from the end of list. Its working well. Butt i don't know about node.next.next?? please help me in understanding of concept of nodes. Here is code in which i implemented node.next.next. public void del_`enter code here`end(){ Node temp = head; while(temp.next.next!=null ){ temp= temp.next; } temp

Has to be O(n) and in-place (space complexity of 1). The code below does work, but is there a simpler or better way? public void invert() { if (this.getHead() == null) return; if (this.getHead().getNext() == null) return; //this method should reverse the order of this linked list in O(n) time Node<E> prevNode = this.getHead().getNext(); Node<E> nextNode = this.getHead().getNext().getNext(); prevNode.setNext(this.getHead()); this.getHead().setNext(nextNode); nextNode = nextNode.getNext(); while (this.getHead().getNext() != null) { this.getHead().getNext().setNext(prevNode); prevNode = this

I just struggled through a simple interview question: Please reverse a singly linked list. While I failed to provide a working answer in time to save the interview, I was able to come up with a solution afterwards. Is my solution correct? How would you analyze this with Big-Oh? Are there more efficient ways to reverse a singly linked list? // reverse a linked list var reverseLinkedList = function(linkedlist) { var node = linkedlist; var previous = null; while(node) { // reverse pointer node.next = previous; // increment previous to current node previous = node; // increment node to next node