[Leecode]Linked List CycleII

Linked List Cycle II My Submissions Question
Total Accepted: 59621 Total Submissions: 189735 Difficulty: Medium
Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

Note: Do not modify the linked list.

Follow up:
Can you solve it without using extra space?

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这是一道让我有一些些忧桑的题目

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        if(!head || !head->next)   return NULL;
        ListNode* f = head;
        ListNode* s = head;
        ListNode* p = head;
        ListNode* m = NULL;
        int flag = 0;
        while(f->next && f->next->next){
            s = s->next;
            f = f->next->next;
            if(s == f){
                m = s;
                flag = 1;
                break;//注意不要忘记加break,
            }
        }
        while(p->next && flag == 1){
            if(m == p)  break;//注意这个语句在的位置，如果放在递增语句后面会在[1,2]（2->next = head）这里报错
            p = p->next;
            m = m->next;
        }
        return m;
    }
};