Given a linked list, determine if it has a cycle in it.
Follow up:
Can you solve it without using extra space?
开始以为这道题只需要注意不使用额外空间即可，于是写了个时间复杂度为O(n^2)暴力搜索算法，如下：
/** * Dumped, Time Limit Exceeded */ class Solution { public: bool hasCycle(ListNode *head) { if (!head || !head->next) return false; ListNode *pre = head; ListNode *cur = head->next; while (cur) { pre = head; while (cur->next != pre && pre != cur) { pre = pre->next; } if (cur->next == pre && pre != cur) return true; cur = cur->next; } return false; } };
可是OJ还是不给通过，原因是Time Limit Exceeded，还是需要把时间复杂度缩小的O(n)才行，于是上网搜各位大神的方法，发现果然很神奇。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) { if (!head || !head->next) return false; ListNode *slow = head; ListNode *fast = head; while (slow && fast) { slow = slow->next; fast = fast->next; if (!fast) return false; fast = fast->next; if (slow == fast) return true; } return false; } };
[LeetCode] Linked List Cycle 单链表中的环
原文：http://www.cnblogs.com/grandyang/p/4137187.html