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Reverse a Linked List

EasyLinked listO(n) time · O(1) space

The problem

Reverse a singly linked list and return the new head.

1 → 2 → 3 → null   becomes   3 → 2 → 1 → null
Stuck? Open one hint

Before you repoint a node, you need to have saved where it was going.

Show the worked solution

Three pointers: prev (starts null), current (starts at head), and a temporary next.

In each step: save next, point current.next back to prev, then shift both prev and current forward. When current is null, prev is the new head.

Draw it on the whiteboard as you speak — this question is as much about explaining pointer movement clearly as about getting it right.

KotlinO(n) time · O(1) space
class ListNode(var value: Int) { var next: ListNode? = null }

fun reverseList(head: ListNode?): ListNode? {
    var prev: ListNode? = null
    var current = head
    while (current != null) {
        val next = current.next
        current.next = prev
        prev = current
        current = next
    }
    return prev
}