Leetcode: Linked List Cycle II

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题目：
Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

思路分析：
和《Leetcode: Linked List Cycle 》一样还是双指针的方法。

一个循环链表如图
slow指针走了S=X+Y
fast指针走了F=X+Y+Z+Y
两个指针相遇。
且有：2S=F，则有X=Z。
所以，从head到环开始的路程＝从相遇到环开始的路程。
所以，当slow和fast相遇了，我们拿slow从头开始走，fast从相遇的地方开始走，两个都走一步，那么再次相遇必定是环的开始节点。

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:    ListNode *detectCycle(ListNode *head)    {        if (!head) return nullptr;        ListNode *slow = head;        ListNode *fast = head;        bool hasCycle = false;        while (fast && fast->next)        {            slow = slow->next;            fast = fast->next->next;            if (slow == fast)            {                hasCycle = true;                break;            }        }        if (hasCycle)        {            slow = head;            while (slow != fast)            {                slow = slow->next;                fast = fast->next;            }            return slow;        }        else return nullptr;    }};

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 ListNode DetectCycle(ListNode head)    {        if (head == null) return null;        ListNode slow = head;        ListNode fast = head;        bool hasCycle = false;        while (fast != null && fast.next != null)        {            slow = slow.next;            fast = fast.next.next;            if (slow == fast)            {                hasCycle = true;                break;            }        }        if (hasCycle)        {            slow = head;            while (slow != fast)            {                slow = slow.next;                fast = fast.next;            }            return slow;        }        else return null;    }}

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