public static void main(String[] args) {
ListNode n1 = new ListNode(1);
ListNode n2 = new ListNode(2);
ListNode n3 = new ListNode(3);
ListNode n4 = new ListNode(4);
ListNode n5 = new ListNode(5);
ListNode n6 = new ListNode(6);
n1.next = n2;
n2.next = n3;
n3.next = n4;
n4.next = n5;
n5.next = n6;
n6.next = n3;
LC_141_DetectCycle inst = new LC_141_DetectCycle();
System.out.println(inst.hasCycle(n1));
}
/**
* 1. get catch point with two points(fast/slow) 2. split the cycle on that point 3. get the
* intersection/joint node
*
* <p>如果允许修改每个node的val，可以用染色的方法 node = head while(node != null) { node.val = 999999;
* if(node.next.val == 999999) return node.next; }
*
* @param head
* @return
*/
public ListNode detectCycle(ListNode head) {
if (head == null || head.next == null) return null;
LC_141_DetectCycle detector = new LC_141_DetectCycle();
ListNode p = detector.getCatchPoint(head);
if (p == null) return null;
// split the cycle on the catch point, then get the intersection node
ListNode head2 = p.next;
p.next = null;
LC_160_GetIntersectionNode inst = new LC_160_GetIntersectionNode();
return inst.getIntersectionNode(head, head2);
// another way is a little magic
// suppose the distance between head and joint point is K, then it can be
// proved that the catch point is also K distance far from the joint.
/*
ListNode p1 = head;
while(p1 != p) {
p1 = p1.next;
p = p.next;
}
return p;
*/
}