Leetcode: Linked List Cycle

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

思路分析：
利用快慢指针slow，fast。 slow指针每次走一步，fast指针每次走两步，倘若存在环，则slow和fast必定在某一时刻相遇。

C++参考代码：

/**
 * 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)
    {
        if (!head) return false;
        ListNode *slow = head;
        ListNode *fast = head->next;
        while (fast && fast->next)
        {
            if (slow == fast) return true;
            slow = slow->next;
            fast = fast->next->next;
        }
        return false;
    }
};

另外一种写法：

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

C#参考代码：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution
    public bool HasCycle(ListNode head)
    {
        if (head == null) return false;
        ListNode slow = head;
        ListNode fast = head;
        while (fast != null && fast.next != null)
        {
            slow = slow.next;
            fast = fast.next.next;
            if (slow == fast) return true;
        }
        return false;
    }
}