leetcode_linked_list_cycle

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难度：Easy

解题思路：两个节点slow、fast，slow每次走一步，fast每次走两步。如果有cycle，它们一定会相遇。

代码如下：

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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    bool hasCycle(ListNode *head) {
        ListNode *dummy = new ListNode(-1);
        dummy->next = head;
        ListNode *slow = dummy;
        ListNode *fast = dummy;
        while(fast -> next != NULL && fast ->next -> next != NULL)
        {
            slow = slow->next;
            fast = fast->next->next;
            if(slow == fast)
            {
                return true;
            }
        }
        return false;
    }
};

代码结果：9ms，超过53.73%