How would you reverse a singly-linked list?

Question

Puu · Accepted Answer

public void reverse() {
		
		Node temp1 = head.next;
		Node temp2 = temp1.next;
				
		head.next = null;
		temp1.next = head; // Make head the last node
		
		while(temp2.next != null){
			Node link = temp2.next;
			temp2.next = temp1;
			
			//Increment the positions
			temp1 = temp2;
			temp2 = link;
			
		}
		
                // Reassign the head
		if(temp2.next == null){
			temp2.next = temp1;
			head = temp2;
		}
		
	}

Daniel · Answer

Recursive anser:
var root:Node
//Assuming there is a variable called root.
func reverse(first:Node)
{
    if first.next == nil {
        return
    }
    let oldRoot = root
    root = first.next!
    first.next = first.next!.next
    root.next = oldRoot
    reverse(first)
}

Anonymous · Answer

+ (Node *)reverse:(Node *)root {
  if (!root) {
    return nil;
  }

  if (!root.nextl) {
    return root;
  }

  Node *newRoot = [Node reverse:root.next];
  root.next.next = root;
  root.next = nil;
  return newRoot;
}

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