判斷鏈表有沒有環，用Floyd Cycle Detection算法，用兩個快慢指針。

class Solution {
public:
    bool hasCycle(ListNode *head) {
        if (!head) return false;
        ListNode *slow, *fast;
        slow=fast=head;
        do{
            if (fast==NULL || fast->next==NULL) return false;
            slow = slow->next;
            fast 
 
= fast->next->next;
        }while(slow!=fast);
        return true;
    }
};

142是141的進階，需要額外判斷環的起點。

詳見 https://leetcode.com/problems/linked-list-cycle-ii/solution/ 中的推導，不過裏面應該是 F=(n-1)(a+b)+b，因此從head和相遇點開始同時走，一定會在循環起點相遇。

class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        
 
if (!head) return NULL;
        ListNode *slow, *fast;
        slow = fast = head;
        do{
            if (!fast || !fast->next) return NULL;
            slow = slow->next;
            fast = fast->next->next;
        }while(slow!=fast);
        
        ListNode *p=head, *q=slow;
        
 
while(p!=q){
            p = p->next;
            q = q->next;
        };
        return p;    
    }
};

LeetCode 141, 142. Linked List Cycle I+II