Given a linked list, determine if it has a cycle in it.
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.
Example 1:
Input: head = [3,2,0,-4], pos = 1
Output: true
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: true
Explanation: There is a cycle in the linked list, where tail connects to the first node.
Example 3:
Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.
Follow up:
Can you solve it using O(1) (i.e. constant) memory?
Accepted
Solution:
使用快慢指针，若两个指针会重合，那么就有环，否则没有
1 class Solution { 2 public: 3 bool hasCycle(ListNode *head) { 4 if (head == nullptr || head->next == nullptr)return false; 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 return true; 13 } 14 return false; 15 } 16 };