Problem · LC 206

Reverse Linked List

Easy
◧ Foundation · pointer manipulation

How to recognize it

A singly linked list, and you need to reverse the direction of all pointers. The signal: "reverse", "flip direction", or showing output as last→...→first. This is the building block for many linked list problems.

The core idea

Maintain three pointers: prev (initially null), curr (at head), next (save curr.next before we change it). Iterate through the list, at each step reversing the pointer from curr→next to curr→prev. Move all three pointers forward one step. After the loop, prev points to the new head.

Why It's Tricky

Reversing a linked list looks simple but is deceptively easy to get wrong. The key is that you're modifying pointers while iterating, so you MUST save the next node before breaking the link. Forgetting this causes you to lose the rest of the list.

The Key Insight

The core move: `curr.next = prev` instead of `curr.next = curr.next.next`. You're flipping the pointer direction, not skipping nodes. Always save `next = curr.next` first.

Step-by-Step Walkthrough

Example: 1→2→3→null

Start: prev=null, curr=1
Step 1: next=2, 1.next=null, prev=1, curr=2
Step 2: next=3, 2.next=1, prev=2, curr=3
Step 3: next=null, 3.next=2, prev=3, curr=null
Loop ends, return prev (which is 3)

Result: 3→2→1→null

time O(n)
space O(1)
reverse_linked_list
def reverseList(head):
    prev, curr = None, head
    while curr:
        next = curr.next        # Save next node
        curr.next = prev        # Reverse the pointer
        prev, curr = curr, next # Move forward
    return prev

Interview gotchas

  • Must save next = curr.next BEFORE changing curr.next, or you lose the rest of the list.
  • Initial prev = null — the old head's next should point to null.
  • Return prev, NOT head — head is now the tail and points to null.
  • Off-by-one: Make sure your loop condition is while curr is not null, not while curr.next.

💡 What to Say in the Interview

  • Say: "I'll use three pointers to reverse the links in one pass."
  • Explain: "At each step, I save the next node, reverse the pointer from curr to prev, then move all pointers forward."
  • Emphasize: "Saving next is critical — without it, I'd lose access to the rest of the list."
  • Complexity: "O(n) time (visit each node once), O(1) space (only three pointers)."

↳Variants they'll pivot to

Solve the base problem, then interviewers escalate. These are the usual next steps — knowing the connection is worth more than memorizing each.