LC.141 | 环形链表 | 链表 | 快慢双指针

输入：链表

输出：判定链表内部是否有环

思路：用快慢双指针遍历链表，如果有链表二者一定会相遇；如果没有环，fast会先到nullptr。

class Solution {
public:
    bool hasCycle(ListNode *head) {
        if (head == nullptr || head->next == nullptr) {
            return false;
        }

        ListNode *slow = head;
        ListNode *fast = head;

        while (fast != nullptr && fast->next != nullptr) {
            slow = slow->next;          // Move slow pointer by one.
            fast = fast->next->next;    // Move fast pointer by two.

            if (slow == fast) {
                return true;
            }
        }

        return false;
    }
};