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Linked List Cycle類似，還是用map。

用時：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<ListNode*, bool> 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 };

?

同時，對于Linked List Cycle中的較優方法，同樣適用于本題。當fast和slow指針相遇時，令設指針slow2 = head，那么slow2和slow一定會在相遇的地方重合。

用時：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 };

轉載于:https://www.cnblogs.com/amazingzoe/p/4518313.html

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