Reverse Linked List

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

Example

Key Pointers

• Reversal is just re-pointing each node's next to its predecessor instead of its successor.
• Track a prev pointer starting at None — that becomes the new head once the loop ends.
• Save the next node before overwriting curr.next, or the rest of the list is lost.

Step-by-Step Analysis

1. Initialize prev = None, curr = head.
2. While curr is not None: save next_node = curr.next.
3. Point curr.next = prev, then advance prev = curr, curr = next_node.
4. When curr is None, prev is the new head — return it.

Complexity

Time: O(n) — Space: O(1)

Code

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def reverseList(head: ListNode | None) -> ListNode | None:
    prev = None
    curr = head
    while curr:
        next_node = curr.next
        curr.next = prev
        prev = curr
        curr = next_node
    return prev

Code Explanation

How the solution above works, piece by piece.

1. prev = None: the reversed part starts empty (the old head will end up pointing to None).
2. next_node = curr.next: save the rest of the list before changing any pointer, or it would be lost.
3. curr.next = prev: flip this node to point backwards.
4. prev = curr, curr = next_node: step both pointers forward.
5. When curr is None, prev is the old tail, which is the new head. O(n) time, O(1) space.

Visual Walkthrough

Running it on head = [1, 2, 3, 4, 5]:

Video Explanation

NeetCode — Reverse Linked List, LeetCode 206