力扣算法——142LinkedListCycleII

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?

Solution:

　　使用快慢指针，若快慢指针能重合上，那就有环

　　然后快指针从头移动，与此同时慢指针一起先后移动，当两个指针再次重合时，则就是环的入口！

 class Solution {
 public:
     ListNode *detectCycle(ListNode *head) {
         if (head == nullptr || head->next == nullptr)return nullptr;
         ListNode *slow, *fast;
         slow = fast = head;
         while (fast && fast->next)
         {
             slow = slow->next;
             fast = fast->next->next;
             if (fast == slow)
                 break;
         }
         if (fast != slow)return nullptr;
         slow = head;
         while (slow != fast)
         {
             slow = slow->next;
             fast = fast->next;
         }
         return fast;
     }
 };