leetcode - 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?

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
struct ListNode
{
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};

//这题是Linked list Cycle的变形题，首先用fast,slow指针，判断是否是Cycle list,
//然后，再将slow，指向head头结点，然后继续遍历，当fast == slow的时候，那么，slow的值就是Cycle的入口点。
class Solution {
public:
ListNode *detectCycle(ListNode *head) {
if(head == NULL || head->next == NULL) return head;
ListNode *fast = head;
ListNode *slow = head;
bool isCycle = false;
while(fast != NULL && slow != NULL)
{
slow = slow->next;
if(fast->next == NULL) return NULL;
fast = fast->next->next;
if(fast == slow)
{
isCycle = true;
break;
}
}
if(!isCycle) return NULL;
slow = head;
while(fast != slow)
{
fast = fast->next;
slow = slow->next;
}
return slow;
}
};