题目描述：

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

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

思路：

设置一个快指针fast,一个慢指针slow。快指针一次移动两个单位，慢指针一次移动一个单位。如图，链表进入环之后是逆时针移动，假设两个指针是在红蓝交界的地方相遇。则快指针走的距离Sf=x+y+z+y=x+2y+z，慢指针走的距离Ss=x+y。又因为Sf=2Ss，则x+2y+z=2x+2y，可得x=z。则可令fast指针重新回到head，两指针依次移动一个单位直到相遇。相遇的点即为环开始的点。

代码：

ListNode * Solution::detectCycle(ListNode * head)
{
    if(head == NULL)
        return NULL;

    if(head->next == head)
        return head;

    ListNode * fast = head;
    ListNode * slow = head;

    while(fast != NULL && fast->next != NULL)
    {
        fast = fast->next->next;
        slow = slow->next;
        if(fast == slow)
            break;
    }

    if(fast == NULL || fast->next == NULL)
        return NULL;
    else
    {
        fast = head;
        while(fast != slow)
        {
            fast = fast->next;
            slow = slow->next;
        }
        return fast;
    }
}

LeetCode:Linked List Cycle II