Linked List Cycle II

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

Note: Do not modify the linked list.

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

Analyse: 

Runtime: 12ms. 

/**
 * 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 || !head->next) return NULL;
        
        ListNode* slow = head, *fast = head, *entry = head;
        while(fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;
            
            if(slow == fast) {
                while(entry != slow) {
                    entry = entry->next;
                    slow = slow->next;
                }
                return slow;
            }
        }
        return NULL;
    }
};