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?

Draw a picture and visualize it:

11111222           
          222
Let's say x is the distance from head of the linked list to the entry node where the cycle begins, a is the distance from entry node where the cycle begins to the node where the slow and fast pointers meet, r is the size of the loop. s is the distance where slow pointer has travelled before slow point and fast pointer have met. L is the length from meet point to entry point.

2*s=s+nr;
s = nr;
x+a=nr=(n-1)*r + r = (n-1)*r + L-x;

x=(n-1)*r + L-x-a;

/**
 * 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 *p1=head;
        ListNode *p2=head;
       
        while(p1!=NULL&&p2!=NULL)
        {
            p1=p1->next;
            p2=p2->next==NULL?NULL:p2->next->next;
           
            if (p1==p2)
            {
                p2=head;
                while(p1!=p2)
                {
                    p1=p1->next;
                    p2=p2->next;
                }
                return p1;
            }
        }
       
        return NULL;
    }
};