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

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

思路:维护两个指针,一快一慢,判断两个指针能否相遇。

 class Solution {
public:
bool hasCycle(ListNode *head) {
if (head == NULL) return false;
ListNode *slow = head;
if (head->next == NULL) return false;
ListNode *fast = head->next;
while (slow != fast)
{
if (slow != NULL)
slow = slow->next;
if (fast != NULL)
fast = fast->next;
if (fast != NULL)
fast = fast->next;
}
return slow != NULL;
}
};

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