To decide if a singly linked list has a cycle and find the start node of cycle is such and classical problem. Computer scientist spent some years before find this standard solution. Like many other problems concerns about linked list, in this problem we also need to pointer: a fastRunner and a slowRunner. The fastRunner moves two steps every time while slowRunner moves one. At last either the two pointers are pointing the same node, or the fastRunner meets the end of the list. In the former case, there must be a cycle in the list. In the latter case, there are no cycle.

To find the start node, after the two pointers meets, we move slowRunner back to the head and moves two pointer one step every time in parallel until they meet again. The node they meet at is the start point of cycle.

Note at first both pointers point to head, at this time we cannot say they meet and we must wait for the second time they are both pointing to a same node.

/**
* class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) {
*         val = x;
*         next = null;
*     }
* }
*/

while (fastRunner != null && fastRunner.next != null) {
fastRunner = fastRunner.next.next;
slowRunner = slowRunner.next;

if (fastRunner == slowRunner) {
// find a cycle