單鏈表,問有沒有環,若有,找到環的入口.

Lintcode上挑戰,只許使用常數的空間.

設一個快指針fast,一個慢指針slow,fast每次走2步,slow每次走1步.

若無相遇找到了鏈尾,則無環,若相遇了,則有環.

設從鏈頭到環入口點走了a步,從環入口到相遇點走了x步,環長r.相遇時,slow走了s步,fast走了2s步.

有:

s=a+x, 2s=a+nr+x.

所以a+x=nr->a=nr-x.

註意理解a=nr-x.意思是從鏈頭走到環入口的 步數=再走n圈,並退x步.那剛好也在環入口.

 1 class Solution {
 2 public:
 3     ListNode *detectCycle(ListNode *head) {
 
 4         ListNode *slow = head, *fast = head;
 5         while (fast && fast->next) {
 6             slow = slow->next;
 7             fast = fast->next->next;
 8             if (slow == fast)
 9                 break;
10         }
11         if (!fast || !fast->next)
12             return
 
 NULL;
13         slow = head;
14         while (slow != fast) {
15             slow = slow->next;
16             fast = fast->next;
17         }
18         return slow;
19     }
20 };

帶環鏈表 II