Reverse Linked List

If there's one Linked List pattern that every developer should master, it's reversing a Linked List . This problem teaches pointer manipulation, which later helps in solving problems like: Reverse Nodes in K Group Palindrome Linked List Reorder List Reverse Between Positions Let's understand the intuition behind it. Problem Statement Given the head of a singly linked list, reverse the list and return the new head. Example Input: 1 -> 2 -> 3 -> 4 -> 5 Output: 5 -> 4 -> 3 -> 2 -> 1 Brute Force Approach Intuition A straightforward approach is to store all node values inside an ArrayList. Then traverse the linked list again and replace node values from the array in reverse order. Although it works, we're using extra space just to remember previous elements. Complexity Time Complexity : O ( N ) Space Complexity : O ( N ) Brute Force Code class Solution { public ListNode reverseList ( ListNode head ) { ArrayList < Integer > values = new ArrayList <>(); ListNode temp = head ; while ( temp != null ) { values . add ( temp . val ); temp = temp . next ; } temp = head ; int index = values . size () - 1 ; while ( temp != null ) { temp . val = values . get ( index --); temp = temp . next ; } return head ; } } Moving Towards the Optimal Approach Instead of storing values separately, can we reverse the actual links between nodes? Since Linked Lists are made of pointers, reversing those pointers directly would save extra space. This leads us to the optimal solution. Optimal Approach - Three Pointer Technique We'll maintain three pointers: prev curr next For every node: Store the next node. Reverse the current node's pointer. Move all pointers one step ahead. Visual Intuition Initially: null <- 1 -> 2 -> 3 -> 4 -> 5 ^ curr After first reversal: null <- 1 2 -> 3 -> 4 -> 5 ^ prev ^ curr Continue the same process until the list ends. Dry Run Input 1 -> 2 -> 3 -> null Initial State prev = null curr = 1 Iteration 1 next = 2 1 -> null prev = 1 curr = 2 Iteration 2 next = 3 2 -> 1 -> null p