leetcode: Linked List Cycle II

http://oj.leetcode.com/problems/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：

如果前面Linked List Cycle的问题已经解决，再扩展一下就可以了。如果一步走两步走的方法在某个点交汇了，那么这个点肯定在环内。那么算出环的长度n，令某个节点从head出发先走n步，然后与head节点一起前进，那么交汇点就是环的起点。

 1 class Solution {
 2 public:
 3     ListNode *detectCycle(ListNode *head) {
 4         if (NULL == head) {
 5             return NULL;
 6         }
 7
 8         ListNode *curr = head, *next = head->next;
 9
10         while (true) {
11             curr = curr->next;
12
13             int steps = 0;
14
15             while ((next != NULL) && (steps < 2)) {
16                 next = next->next;
17                 ++steps;
18             }
19
20             if (NULL == next) {
21                 return NULL;
22             }
23
24             if (curr == next) {
25                 break;
26             }
27         }
28
29         ListNode *next_head = head->next;
30
31         next = next->next;
32
33         while (curr != next) {
34             next = next->next;
35             next_head = next_head->next;
36         }
37
38         while (head != next_head) {
39             head = head->next;
40             next_head = next_head->next;
41         }
42
43         return head;
44     }
45 };

转载于:https://www.cnblogs.com/panda_lin/p/linked_list_cycle_ii.html

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