#206 - Reverse Linked List

EasyLinked List

Given the head of a singly linked list, reverse the list and return the new head.

Input	Output
head = [1,2,3,4,5]	[5,4,3,2,1]
head = [1,2]	[2,1]
head = []	[]

Solution Space

Try it

/**
 * Reverses a singly linked list in-place and returns the new head.
 *
 * @param head Head of the original list (may be null)
 * @returns    Head of the reversed list
 */
export function reverseList(head: ListNode | null): ListNode | null {
  let prev: ListNode | null = null;
  let curr = head;

  while (curr) {
    const next = curr.next; // save next before overwriting
    curr.next = prev;
    prev = curr;
    curr = next;
  }

  return prev;
}

Approach: iterative pointer reversal – maintain three pointers (`prev`, `curr`, `next`). On each step, point `curr.next` back to `prev`, then advance all three.

Time: O(n) – single pass through the list.

Space: O(1) – only three pointer variables.

Alternative approaches:

Approach	Idea	Time	Space
Recursive	Base case: single node or null. Recurse on head.next, then set head.next.next = head and head.next = null.	O(n)	O(n) call stack

Why the iterative method is preferred: identical time complexity with O(1) space instead of O(n) stack frames, and no risk of stack overflow on very long lists.

#141 Linked List Cycle

#328 Odd Even Linked List