题解 | #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) {}
 * };
 */
#include <cmath>
class Solution {
public:
    bool hasCycle(ListNode *head) {
        if (head == nullptr) {
            return false;
        }
        
        /* 如果不破坏链表的话，设置两个步长不同的指针，快的指针会把慢的套圈 */
        ListNode* slowPtr = head;
        ListNode* fastPtr = head->next;

        while (fastPtr != nullptr && fastPtr->next != nullptr) {
            if (fastPtr == slowPtr) {
                return true;
            }
            fastPtr = fastPtr->next->next;
            slowPtr = slowPtr->next;
        }
        return false;

    }
};