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 head while(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; }
时间: 2024-10-12 20:14:45