Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
Note: Do not modify the linked list.
Follow up:
Can you solve it without using extra space?
题意：141的followup，寻找环的起点。
思路：在141求到两个快慢指针的相遇节点meet后，再用一个指针dummy指向head，然后让head和meet同时各向前走一步，最后他们的交点就是环的起点。
推导过程见链接，http://www.cnblogs.com/hiddenfox/p/3408931.html。
public ListNode detectCycle(ListNode head) {
if (head == null) {
return null;
}
ListNode meet = null;
ListNode slow = head;
ListNode fast = head;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
if (slow == fast) {
meet = slow;
break;
}
}
if (meet == null) {
return null;
}
ListNode dummy = head;
while (dummy != meet) {
dummy = dummy.next;
meet = meet.next;
}
return dummy;
}