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 poswhich 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 = [,,,-], pos =
Output: tail connects to node index
Explanation: There is a cycle in the linked list, where tail connects to the second node.
Example 2:
Input: head = [,], pos =
Output: tail connects to node index
Explanation: There is a cycle in the linked list, where tail connects to the first node.
Example 3:
Input: head = [], pos = -
Output: no cycle
Explanation: There is no cycle in the linked list.
这题解题的思路在于：在第一次相遇点位置pos，从该位置到环入口Join的距离=从头结点Head到环入口Join的距离
假设环的长度是r，在第一次相遇时，慢指针走过的路程：s=lenA+x，快指针走过的路程：2s=lenA+nr+x，所以：lenA+x=nr，即：LenA=nr-x。
所以在第一次相遇之后，一个指针从head走到join的路程，另一个指针从pos走到join。
方法一（C++）
ListNode *detectCycle(ListNode *head) {
ListNode* slow=head,*fast=head;
while(fast&&fast->next){
slow=slow->next;
fast=fast->next->next;
if(slow==fast)
break;
}
if(!fast||!fast->next)
return NULL;
slow=head;
while(slow!=fast){
slow=slow->next;
fast=fast->next;
}
return slow;
}
（java）：
ListNode slow=head,fast=head;
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;
slow=head;
while(slow!=fast){
slow=slow.next;
fast=fast.next;
}
return slow;
}