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?
参考http://www.cnblogs.com/hiddenfox/p/3408931.html
方法：
第一次相遇时slow走过的距离：a+b，fast走过的距离：a+b+c+b。
因为fast的速度是slow的两倍，所以fast走的距离是slow的两倍，有 2(a+b) = a+b+c+b，可以得到a=c（这个结论很重要！）。
我们发现L=b+c=a+b，也就是说，从一开始到二者第一次相遇，循环的次数就等于环的长度。
我们已经得到了结论a=c，那么让两个指针分别从X和Z开始走，每次走一步，那么正好会在Y相遇！也就是环的第一个节点。
/**
* Definition for singly-linked list.
* class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode detectCycle(ListNode head) {
if(head == null || head.next == null){
return null;
} ListNode fast = head, slow = head; while(1 == 1){
if(fast == null || fast.next ==null) return null; fast = fast.next.next;
slow = slow.next;
if(fast == slow){
break;
}
} slow = head;
while(fast != slow){
fast = fast.next;
slow = slow.next;
} return slow; }
}
或者
/**
* Definition for singly-linked list.
* class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode detectCycle(ListNode head) {
if(head == null || head.next == null){
return null;
} ListNode fast = head.next, slow = head; while(fast != slow){
if(fast == null || fast.next ==null) return null; fast = fast.next.next;
slow = slow.next;
} while(head != slow.next){
head = head.next;
slow = slow.next;
} return head; }
}