141. Linked List Cycle (List; Two-Pointers)

Given a linked list, determine if it has a cycle in it.

Follow up:
Can you solve it without using extra space?

思路：采用“快慢指针”查检查链表是否含有环。让一个指针一次走一步，另一个一次走两步，如果链表中含有环，快的指针会再次和慢的指针相遇。

class Solution {
public:
    bool hasCycle(ListNode *head) {
        ListNode* slow = head;
        ListNode* fast = head;  

        while(fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;
            if(slow == fast)
                return true;
        }
        return false;
    }
};