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这道题是快慢指针的经典应用。只需要设两个指针,一个每次走一步的慢指针和一个每次走两步的快指针,如果链表里有环的话,两个指针最终肯定会相遇。实在是太巧妙了,要是我肯定想不出来。代码如下:
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public: bool hasCycle(ListNode *head) { ListNode *slow = head, *fast = head; while(fast && fast->next) { slow = slow->next; fast = fast->next->next; if(slow==fast) return true; } return false; }};
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https://www.cnblogs.com/grandyang/p/4137187.html