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?

分析：和类似，还是用map。

用时：60ms

1 /** 2  * Definition for singly-linked list. 3  * struct ListNode { 4  *     int val; 5  *     ListNode *next; 6  *     ListNode(int x) : val(x), next(NULL) {} 7  * }; 8  */ 9 class Solution {10 public:11     ListNode *detectCycle(ListNode *head) {12         map
     
       m;13         while(head){14             if(m.find(head) == m.end()) m[head] = true;15             else return head;16             head = head->next;17         }18         return NULL;19     }20 };
     

 

同时，对于中的较优方法，同样适用于本题。当fast和slow指针相遇时，令设指针slow2 = head，那么slow2和slow一定会在相遇的地方重合。

用时：16ms

1 /** 2  * Definition for singly-linked list. 3  * struct ListNode { 4  *     int val; 5  *     ListNode *next; 6  *     ListNode(int x) : val(x), next(NULL) {} 7  * }; 8  */ 9 class Solution {10 public:11     ListNode *detectCycle(ListNode *head) {12         ListNode *fast = head, *slow = head;13         while(fast && fast->next){14             fast = fast->next->next;15             slow = slow->next;16             if(fast == slow){17                 ListNode* slow2 = head;18                 while(slow){19                     if(slow2 == slow) return slow;20                     slow = slow->next;21                     slow2 = slow2->next;22                     23                 }24             }25         }26         return NULL;27     }28 };