141. Linked List Cycle

Given a singly linked list, determine if it is a palindrome.

Follow up:
Could you do it in O(n) time and O(1) space?

快慢指針

1. 快指針從head+1出發，慢指針從head出發， 當快能夠追上慢，說明有環
2. 需判斷頭結點/頭指針為空的清情形

 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     bool hasCycle(ListNode *head) {
12         if (head == nullptr || head->next == nullptr)
13             return 0;
14         
15         ListNode* pSlow = head;
16         ListNode* pFast = head->next;
 
17         
18         while (pFast != nullptr && pSlow != nullptr)
19         {
20             if (pFast == pSlow)
21                 return 1;
22             
23             pSlow = pSlow->next;
24             pFast = pFast->next;
25             
26             if (pFast != nullptr)
27
 
                 pFast = pFast->next;
28         }
29         return 0;
30     }
31 };

141. Linked List Cycle 判斷鏈表中是否存在“環”