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Problem: 24. Swap Nodes in Pairs To swap two adjacent nodes, we need a pointer standing just before the pair. A dummy sentinel node placed before the head simplifies edge cases. The traversal pointer curr starts at the sentinel. Swapping involves rerouting next pointers in three steps while preserving references that might be lost. The loop con ...

Pairwise Node Swapping in Linked List Given a linked list, swap every two adjacnet nodes and return the modified list's head. Node values must not be altered; only node positions can be chenged. Example: Input: head = [1,2,3,4] Output: [2,1,4,3] Solution: Use three pointers to manage node connections during swapping. class Solution { publi ...

Problem DescriptionGiven the head of a linked list, determine the node where a cycle begins. If no cycle exists, return null. The cycle is identified when a node can be reached again by continuously following the next pointer. The solution must not modify the original linked list.Algorithm ExplanationFloyd's Cycle Detection Algorithm, also know ...

Swapping Nodes in Pairs Problem: Given a linked list, swap every two adjacent nodes and return its head. Iterative Approach: class ListNode { int val; ListNode next; ListNode(int val) { this.val = val; } } class Solution { public ListNode swapPairs(ListNode head) { if (head == null || head.next == null) { ...

Swapping Nodes in PairsApproach: Using a dummy head nodeLogic: Create a dummy head node to simplify the swapping process. Use a current pointer that moves forward two steps at a time. The loop continues as long as there are at least two more nodes to swap.Implementation:/** * Definition for singly-linked list. * struct ListNode { * int v ...