Linked List Cycle I & II

题目：

Given a linked list, determine if it has a cycle in it.

Follow up:
Can you solve it without using extra space?

思路： 

We can use two pointers: walker and runner. walker move one step at a time. runner move two steps. If there is a cycle, walker and runner will keep running in the cycle. And runner will catch up walker at last.

    public boolean hasCycle(ListNode head) {
        if (head == null || head.next == null) {
            return false;
        }
        ListNode walker = head;
        ListNode runner = head.next;
        while (runner != null && runner.next != null) {
            if (walker.val == runner.val) {
                return true;
            }
            walker = walker.next;
            runner = runner.next.next;
        }
        return false;
    }

 For Linked List Cycle II, I have a solution, but have no idea why it is not working....Any idea?

    public ListNode detectCycle(ListNode head) {
        if (head == null) {
            return null;
        }
        ListNode walker = head;
        ListNode runner = head;
        ListNode meet = null;
        
        while (runner != null && runner.next != null) {
            walker = walker.next;
            runner = runner.next.next;
            
            if (runner == walker) {
                meet = walker;
                break;
            }
        }
        
        if (meet == null) {
            return null;
        }
        
        walker = head;
        while (true) {
            if (runner == walker) {
                return walker;
            }
            runner = runner.next;
            walker = walker.next;
        }
    }