Nagibator & Memo
Nagibator
Yo Memo, just smashed my own record on a tough LeetCode marathon—got 5 stars in 30 minutes. Think you can out-code me? Let's see if your precision can keep up with my speed.
Memo
Nice record, that’s impressive. I’m more about clean, bug‑free code than raw speed, but I’m up for a friendly showdown—send me a problem and let’s see who can hit that 5‑star mark first.
Nagibator
Alright, here’s a classic: Reverse a singly linked list in place and return the new head. 5-star, no extra space, no dummy nodes. Who’s gonna finish first? Let the battle begin!
Memo
Sure thing, here’s the classic in‑place reversal, no dummy nodes, no extra space, just three pointers: prev, curr, next. The idea is to iterate through the list, rewire each node’s next to point back to its predecessor, and finally return the last node processed as the new head.
```
Node reverseLinkedList(Node head) {
Node prev = null
Node curr = head
while curr != null {
Node next = curr.next // remember the next node
curr.next = prev // reverse the link
prev = curr // move prev forward
curr = next // move curr forward
}
return prev // new head of the reversed list
}
```
That’s it—O(n) time, O(1) space. Let me know how you’d optimize or if you’re looking for a recursive version. Good luck!
Nagibator
Nice clean code, but I’d give you a twist: handle null or single-node lists in a single line and throw in a tail-call optimization. Think you can beat that? Let’s see what you’ve got.
Memo
Node reverse(Node head) { return (head==null||head.next==null)?head:reverseTail(head,null); } Node reverseTail(Node curr, Node prev) { return curr==null?prev:reverseTail(curr.next,curr); }
Nagibator
Nice recursion, but you just gave me a 5‑star cheat sheet—no speed, no extra space, no dummy nodes, just pure elegance. I’ll keep that in mind for next time. Ready to see who actually crushes a leaderboard? Let's do this.