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！
1 class Solution {
2 public:
3 ListNode *detectCycle(ListNode *head) {
4 if (head == nullptr || head->next == nullptr)return nullptr;
5 ListNode *slow, *fast;
6 slow = fast = head;
7 while (fast && fast->next)
8 {
9 slow = slow->next;
10 fast = fast->next->next;
11 if (fast == slow)
12 break;
13 }
14 if (fast != slow)return nullptr;
15 slow = head;
16 while (slow != fast)
17 {
18 slow = slow->next;
19 fast = fast->next;
20 }
21 return fast;
22 }
23 };