Freaks City

Reversing a Linnked List Problem: Given the head of a singly linked list, reverse the list and return the new head. Approach: Iterative Node Reversal To reverse a linked list iteratively, we can utilize three pointers: current, previous, and temporary. The current pointer traverses the list, while the previous pointer keeps track of the reverse ...

Funtcion Interface Definition typedef int ElementType; typedef struct _dnode { ElementType value; struct _dnode *previous; struct _dnode *next; } DNode; typedef DNode* DList; DList initializeList(); void appendNode(DList list, ElementType value); bool isEmpty(DList list); void forwardTraverse(DList list); void backwardTraverse(DL ...

Pairwise Node Swapping To swap adjacent nodes in pairs, we utilize a dummy node to simplify edge cases. The core idea involves manipulating pointers to reverse each pair while maintaining proper linkage with the rest of the list. A cursor pointer tracks the predecessor of each pair being processed. The termination condition varies based on whet ...

Algorithm Overview Split: Use slow-fast pointer technique to locate the midpoint and partition the list into two halves. Recurse: Apply the same sorting procedure recursively on both halves. Merge: Combine the two sorted sublists into a single sorted list using a linear-time merge step. Implementation class ListNode { int value; ListN ...

Problem Description Given the head of a singly linked list, determine if the list is a palindrome. Return true if it is, otherwise return false. An optimal solution achieves O(n) time complexity and O(1) space complexity by combining a fast-slow pointer approach to find the middle node with a reversal of the latter half of the list. Algorithm O ...

Sentinel Node Strategy A dummy or sentinel node simplifies edge cases where the head pointer might change. By initializing a new node that points to the original head, operations such as deletion or swapping can be treated uniformly with out special casing the start of the list. auto* sentry = new ListNode(0); sentry->next = head; 24. Swap ...

A linked list is a linear data structure where elements, called nodes, are connected via pointers. Each node contains two parts: a data field and a pointer field that references the next node in the seqeunce. The last node's pointer is null, indicating the end of the list. The first node is known as the head. Types of Linked Lists Singly Linked ...

24. Swap Nodes in Pairs Problem Link: LeetCode 24 Solution: This problem involves swapping adjacent nodes in a linked list in pairs. A dummy head node simplifies edge case handling, such as when the list has only one or two nodes. struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(nullptr) {} }; class Soluti ...

Pairwise Swapping of Adjacetn Nodes in a Linked List Given a linked list, swap every two adjacent nodes and return the head of the modified list. The operation must be performed by exchanging nodes, not by altering their internal values. Implementation approach: Use a dummy head node to simplify edge cases. Iterate through the list, adjusting p ...

Given two singly-linked lists that are already sorted in ascending order, produce a single sorted linked list that interleaves all node from both input lists. Examples Input: list1 = [1, 2, 4], list2 = [1, 3, 4] Output: [1, 1, 2, 3, 4, 4] Input: list1 = [], list2 = [] Output: [] Input: list1 = [], list2 = [0] Output: [0] Implementation strate ...