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Given a linked list, return the node where the cycle begins. If there is no cycle, return null
.
Note: Do not modify the linked list.
Follow up:
Can you solve it without using extra space?
Analyse:
Runtime: 12ms.
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode *detectCycle(ListNode *head) { if(!head || !head->next) return NULL; ListNode* slow = head, *fast = head, *entry = head; while(fast && fast->next) { slow = slow->next; fast = fast->next->next; if(slow == fast) { while(entry != slow) { entry = entry->next; slow = slow->next; } return slow; } } return NULL; } };
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原文地址:http://www.cnblogs.com/amazingzoe/p/5743224.html