题解 | #快慢指针-判断链表中是否有环#

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
 /*快慢指针，若有环，必然会再次相遇，否则快指针会跑到NULL*/
class Solution {
public:
    bool hasCycle(ListNode *head) {
        if(head == nullptr) return false;
        ListNode *slow = head;
        ListNode *fast = head;
        while(fast !=nullptr && fast->next != nullptr){
            fast = fast->next->next; //快指针每次跑两步
            slow = slow->next;//慢指针每次跑一步
            if(slow == fast) return true;
        }
        return false;
    }
};