2.8 Loop Detection

For this problem, we have to do a little math deduction:
let L1 indicates the distance between start point and head; L2 indicates the distance between start point and meet point; C indicates the length of cycle;
For fast pointer, it goes L1 + n*C + L2 before meet with slow pointer. And for slow pointer, it goes L1 + L2. We know the speed relationship between slow and fast is that

$$fast = 2*slow$$ Thus we have $$L1 = n*C - L2$$ Which gives us a conclusion: the distance between head and start equals to the distance between meet point to start point along forward movement.

    ListNode *detectCycle(ListNode *head) {
        // write your code here
        if(!head || !head->next) return NULL;
        ListNode *fast = head;
        ListNode *slow = head;
        ListNode *start = head;
        while(fast->next && fast->next->next){
            fast = fast->next->next;
            slow = slow->next;
            if(fast == slow){
                while (slow != start){
                    slow = slow->next;
                    start = start->next;
                }
                return start;
             }
        }
        return NULL;
    }