LeetCode OJ-- Linked List Cycle II **

https://oj.leetcode.com/problems/linked-list-cycle-ii/

判断一个链表中是否有环，如果有，求环的开始位置。

按照上道题目的想法，先判断出是否有环来，同时能得出 slow走了多少步设为 paces。也就是说再次相遇的时候，fast比slow多走了环的n倍的大小。

然后slow = head, fast = head,然后让fast = fast->next往后走 paces 步，之后slow 和 fast再一起走，等到相遇的时候，就是那个环开始地方。

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;  

 struct ListNode {
     int val;
    ListNode *next;
     ListNode(int x) : val(x), next(NULL) {}
};

class Solution {
public:
    ListNode *detectCycle(ListNode *head)  {
        if(head == NULL)
            return false;

        ListNode *slow = head;
        ListNode *fast = head;

        int space = ;
        while(fast)
        {
            if(fast->next)
                fast = fast->next->next;
            else
                return NULL;

            slow = slow->next;

            space++;

            if(fast == slow)
                break;
        }

        if(fast == NULL)
            return NULL;
        slow = head;
        fast = head;
        while(space--)
        {
            fast = fast->next;
        }
        while(fast!=slow)
        {
            fast = fast->next;
            slow = slow->next;
        }
        return slow;
    }
};
int main()
{
    class Solution mys;
    ListNode *n1 = new ListNode();
    ListNode *n2 = new ListNode();
    /*ListNode *n3 = new ListNode(3);
    ListNode *n4 = new ListNode(4);*/
    n1->next = n2;
    /*n2->next = n3;
    n3->next = n4;
    n4->next = n3;*/
    mys.detectCycle(n1);
}