算法描述：

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 posis -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?

解题思路：首先用快慢两个指针判断是否有环。如果有环，再设一额外指针从头开始，以相同的速度移动，直到与慢指针相遇。此时慢指针指向的位置为环所在位置。注意边界值的问题。

    ListNode *detectCycle(ListNode *head) {
        if(head == nullptr || head->next == nullptr) return nullptr;
        ListNode* fast = head;
        ListNode* slow = head;
        bool hasCycle = false;
        while(fast->next!=nullptr && fast->next->next!=nullptr){
            fast=fast->next->next;
            slow=slow->next;
            if(fast == slow){
                hasCycle = true;
                break;
            } 
        }
        if(!hasCycle) return nullptr;
        fast = head;
        while(fast!=slow){
            fast=fast->next;
            slow=slow->next;
        }
        return fast;
    }