LeetCode 206: Reverse Linked List – (Easy)

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

Approach & Solution

We reverse a linked list by keeping track of two pointers: prev (the previous node) and curr (the current node). At each step,…


This content originally appeared on DEV Community and was authored by Grant Riordan

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

example of reversal

Approach & Solution

We reverse a linked list by keeping track of two pointers: prev (the previous node) and curr (the current node). At each step, we:

  • Save the next node so we don’t lose track of the list.

  • Reverse the current node’s pointer to point to prev.

  • Move prev forward to curr.

  • Move curr forward to the saved next node.

We continue until curr is null, meaning we’ve processed all nodes. At that point, prev will be the new head of the reversed list.

Code

ListNode ReverseList(ListNode head)
{
ListNode prev = null;
ListNode curr = head;

while (curr != null)
{
    ListNode nextNode = curr.next;
    curr.next = prev;
    prev = curr;
    curr = nextNode;
}

return prev; // new head

}


This content originally appeared on DEV Community and was authored by Grant Riordan


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