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。

class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        if (NULL == head) {
            return NULL;
        }
        
        ListNode *curr = head, *next = head->next;
        
        while (true) {
            curr = curr->next;
            
            int steps = 0;
            
            while ((next != NULL) && (steps < 2)) {
                next = next->next;
                ++steps;
            }
            
            if (NULL == next) {
                return NULL;
            }
            
            if (curr == next) {
                break;
            }
        }

        ListNode *next_head = head->next;
        
        next = next->next;
        
        while (curr != next) {
            next = next->next;
            next_head = next_head->next;
        }
        
        while (head != next_head) {
            head = head->next;
            next_head = next_head->next;
        }
        
        return head;
    }
};