[LeetCode] Linked List Cycle II

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?

 

思路:快慢指针的应用。设fast为快指针,每次步进两个节点,慢指针为slow,每次步进一个节点。设环长度为Y,环入口距离起点的距离为X,。两个指针在环内距离入口的第k个节点相遇。当两个指针相遇时,慢指针共走了 S步(S=X+nY+K ),快指针共走了2S步(2S= X+mY+K)步。所以

2×(X+nY+K) = X+mY+K 

=> (m-2n)Y = X+K

=> X = (m-2n-1)Y+(Y-K)

所以另慢指针从起点开始走,每次步进一个节点。快指针从相遇的节点开始每次步进一个节点。则两个指针最终会在环入口处相遇。

如果m-2n-1=0,那么Y-K=X,快指针直接达到入口处。如果不为零,那么快指针将要在圈内转上几个来回才能在入口处与满指针相遇。

无论上述哪个情况,两个指针都走了X步。

时间复杂度O(n),空间复杂度O(1)

 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         if (head == NULL) return head;
13         ListNode *fast = head;
14         ListNode *slow = head;
15         
16         while (fast != NULL && fast->next != NULL) {
17             fast = fast->next->next;
18             slow = slow->next;
19             if (fast == slow) break;
20         }
21         
22         if (fast == NULL || fast->next == NULL) 
23             return NULL;
24             
25         slow = head;
26         while (slow != fast) {
27             slow = slow->next;
28             fast = fast->next;
29         }
30         return slow;
31     }
32 };

 

posted @ 2014-11-04 09:57  vincently  阅读(181)  评论(0编辑  收藏  举报