求链表中环的入口

链表中没环就返回NULL

有就返回环的入口

三种基本思路：

1、快慢指针找到环内的一个node，然后从链表头开始，对于每一个node，看它在不在环中

2、用map存一下访问过的节点地址，看当前node的地址是否在map中

3、其实，经过计算，对于1中，快慢指针相遇的地方，再开始以慢指针开始走，

另一方面，在链表的头部也用一个慢指针开始走，二者相遇的地方便是环的入口

（代码并未进行运行验证）

typedef struct node{int data;struct node * next;}listNode;

//find the first node in the cycle//1.step into the circle first and then for every node, take a loop to make sure//2.store the previous node and compare with the cunrrent node (what structure to store?)//3.after computation,while using slow and fast pointer,// we can get that a slow pointer at the begining and another one // at the encounter position will meet at the entrance of the cycle listNode *findFirstNodeInCycle1(listNode *pHead){listNode *pFast=pHead;listNode *pSlow=pHead;while(pFast!=NULL&&pFast->next!=NULL){pFast=pFast->next->next;pSlow=pSlow->next;if(pSlow==pFast)break;}if(pFast==NULL||pFast->next==NULL)return NULL;//now the nodes are in the loop//begin with the headwhile(pHead){pSlow=pSlow->next;while(pSlow){if(pSlow==pHead)return pHead;if(pSlow==pFast)break;pSlow=pSlow->next;}pHead=pHead->next;}}//store in a map? good or not?listNode *findFirstNodeInCycle2(listNode *pHead){if(pHead==NULL)return;listNode *temp=pHead-next;map<int,char> storeMap;map[int(pHead)]=' ';while(teamp!=NULL&&storeMap.find(temp)==storeMap.end()){storeMap[int(temp)]=' ';temp=temp->next;}return temp;}listNode *findFirstNodeInCycle3(listNode *pHead){listNode *pFast=pHead;listNode *pSlow=pHead;while(pFast!=NULL&&pFast->next!=NULL){pFast=pFast->next->next;pSlow=pSlow->next;if(pFast==pSlow){listNode *pSlow2=pHead;while(pSlow2!=pSlow){pSLow=pSlow->next;pSlow2=pSlow2->next;}return pSlow;}}return NULL;}