Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.
Note: Do not modify the linked list.
Example 1:
Input: head = [3,2,0,-4], pos = 1 Output: tail connects to node index 1 Explanation: There is a cycle in the linked list, where tail connects to the second node.
Example 2:
Input: head = [1,2], pos = 0 Output: tail connects to node index 0 Explanation: There is a cycle in the linked list, where tail connects to the first node.
Example 3:
Input: head = [1], pos = -1 Output: no cycle Explanation: There is no cycle in the linked list.
Follow up:
Can you solve it without using extra space?
题目大意：
如果链表中存在环，返回环的入口，否则返回空。
解法：
1.先使用快慢指针，快指针走两步，慢指针走一步，判断链表中是否存在环，如果快慢指针相遇，则不存在，否则快指针先走到链表最后，返回null。
2.然后再让快慢指针相遇一次，计算得到环的长度
3.重新让快慢指针都指向头结点，快指针先走环的长度的步数，当快慢指针再相遇的时候，该节点就是环的入口。
java:
public class Solution {
public ListNode detectCycle(ListNode head) {
ListNode slow=head,fast=head;
if(head==null) return null;
while(fast!=null&&fast.next!=null){
slow=slow.next;
fast=fast.next.next;
if(slow==fast) break;
}
if(fast==null || fast.next==null) return null;
int count=0;
while(count==0||slow!=fast){
slow=slow.next;
fast=fast.next.next;
count++;
}
slow=head;
fast=head;
while(count!=0){
fast=fast.next;
count--;
}
while(slow!=fast){
slow=slow.next;
fast=fast.next;
}
return slow;
}
}