Given a linked list, return the node where the cycle begins. If there is no
cycle, return null.
Follow up:
Can you solve it without using extra space?
老题，没难度。
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/** * 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) { ListNode *fast,*slow; if(head == NULL||head->next==NULL||head->next->next==NULL)return
NULL; fast =head->next->next; slow =head->next; if(fast == NULL ||slow == NULL)return
NULL; while(fast!=slow) { if(fast == NULL ||slow == NULL)return
NULL; fast = fast->next; if(fast == NULL)return
NULL; fast = fast->next; slow = slow->next; } fast = head; while(fast!=slow) { fast=fast->next; slow=slow->next; } return
slow; }}; |
Linked List Cycle II,布布扣,bubuko.com
原文：http://www.cnblogs.com/pengyu2003/p/3625902.html