Linked List Cycle II

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?

分析：

假設有環，鏈表分兩部分，環外部分和環內部分，環外部分長爲L（是邊的長度，不是點的個數），環內長爲S，設置快慢指針，相遇點在環內距離起始點T處，那麼快指針和慢指針之間有如下關係：L+m*S+T = 2*(L+n*S+T), 化簡得(m-2*n-1)*S+(S-T) = L， 那麼一個指針從鏈表起始點出發，另一個指針從相遇點出發，二者會在環的起始點相遇。

/**
 * 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 == NULL||head->next == NULL)
            return NULL;
        ListNode *fast = head, *slow = head;
        while(fast != NULL)
        {
            fast = fast->next;
            if(fast!=NULL)fast = fast->next;
            else
                return NULL;
            slow = slow->next;
            if(fast == slow)
                break;
        }
        if(fast == NULL)
            return NULL;
        ListNode *p = head;
        ListNode *q = fast;
        while(p!=q)
        {
            p = p->next;
            q = q->next;
        }
        return p;
    }
};