链表中的入口节点

题目：如果一个链表中包含环，如何找到环的入口节点？例如，在下图1所示的链表中，环的入口节点是节点3。

图1 链表中环的入口节点

【解题思路】
1、确定一个链表是否包含环。
定义两个指针（即pslow和pfast），同时从链表出发，pfast每次走两步，pslow每次走一步。如果无环，那么pfast将走到链尾而没有与pslow相遇；如果有环，那么pfast将与slow相遇。

2、找到环的入口节点。
先定义两个指针（即pahead和pbehind）。假设链表中的环有n个节点，先让phead指针走n步，再让pslow与pahead同时走，当它们相遇时，相遇点就是入口节点。
如何求链表中环的节点n？
pfast与pslow的相遇节点必在环内。先定义一个指针（即listloopnode）并初始化指向环的入口节点和一个计数变量（即loopnodecount），一边listloopnode向前走，一边loopnodecount计数，当再次回到环入口节点处时，loopnodecount就可以得到环的节点数n。

【代码】
代码1：

Node *MeetingNode(pList pnode){    assert(pnode);    pNode pslow = pnode->next;    if (pslow == NULL)    {        return NULL;    }    pNode pfast = pslow->next;    while (pfast != NULL && pslow != NULL)    {        if (pfast == pslow)        {            return pfast;        }        pslow = pslow->next;        pfast = pfast->next;        if (pfast != NULL)        {            pfast = pfast->next;        }    }    return NULL;}Node* EnterNodeOfLoop(pList pnode){    pNode meetingnode = MeetingNode(pnode);    if (meetingnode == NULL)    {        return NULL;    }    //得到环中节点的数目    int loopnodecount = 1;    pNode pahead = meetingnode;    while (pahead->next != meetingnode)    {        pahead = pahead->next;        ++loopnodecount;    }    //先移动phead，次数为环中的节点数目meetingcount    pahead = pnode;    for (int i = 0; i < loopnodecount; i++)    {        pahead = pahead->next;    }    //再同时移动phead和pbehind    pNode pbehind = pnode;    while (pahead != pbehind)    {        pahead = pahead->next;        pbehind = pbehind->next;    }    return pahead;}

代码2：

//判断链表带环情况pNode CheckCircleEnterNode(pList plist){    pNode fast = plist;    pNode slow = plist;    while (fast && fast->next)    {        fast = fast->next->next;        slow = slow->next;        if (fast == slow)//带环        {            return fast;        }    }    return NULL;}//求环的长度int GetCircleLength(pNode meet){    pNode pnode = meet->next;//meet为相遇点    int count = 1;    while (pnode != meet)    {        pnode = pnode->next;        count++;    }    return count;}//求环的入口点pNode GetCycleEntryNode(pList plist, pNode meet){    pNode pnode = plist;    if ((plist == NULL) && (meet == NULL))    {        return NULL;    }    while (pnode != meet)    {        meet = meet->next;        pnode pnode->next;    }    return meet;}

【程序】

#define _CRT_SECURE_NO_WARNINGS 1# include <stdio.h># include <stdlib.h># include <assert.h>typedef int DataType;typedef struct Node{    DataType data;    struct Node *next;}Node, *pList, *pNode;void InitLinkList(pList *pplist){    assert(pplist);    *pplist = NULL;}void PushBack(pList *pplist, DataType x){    pNode cur = *pplist;    pNode pnode = NULL;    assert(pplist);    pnode = (pNode)malloc(sizeof(Node));    if (pnode == NULL)    {        perror("PushBack::malloc");        exit(EXIT_FAILURE);    }    pnode->data = x;    pnode->next = NULL;    if (cur == NULL)    {        *pplist = pnode;    }    else    {        while (cur->next != NULL)        {            cur = cur->next;        }        cur->next = pnode;    }}void Display(pList plist){    pNode cur = plist;    while (cur)    {        printf("%d-->", cur->data);        cur = cur->next;    }    printf("over\n");}void PrintNode(pList plist){    pNode cur = plist;    printf("%d\n", cur->data);}void Destroy(pList *pplist){    pNode cur = *pplist;    assert(pplist);    while (cur)    {        pNode del = cur;        cur = cur->next;        free(del);    }    *pplist = NULL;}Node *MeetingNode(pList pnode){    assert(pnode);    pNode pslow = pnode->next;    if (pslow == NULL)    {        return NULL;    }    pNode pfast = pslow->next;    while (pfast != NULL && pslow != NULL)    {        if (pfast == pslow)        {            return pfast;        }        pslow = pslow->next;        pfast = pfast->next;        if (pfast != NULL)        {            pfast = pfast->next;        }    }    return NULL;}Node* EnterNodeOfLoop(pList pnode){    pNode meetingnode = MeetingNode(pnode);    if (meetingnode == NULL)    {        return NULL;    }    //得到环中节点的数目    int loopnodecount = 1;    pNode pahead = meetingnode;    while (pahead->next != meetingnode)    {        pahead = pahead->next;        ++loopnodecount;    }    //先移动phead，次数为环中的节点数目meetingcount    pahead = pnode;    for (int i = 0; i < loopnodecount; i++)    {        pahead = pahead->next;    }    //再同时移动phead和pbehind    pNode pbehind = pnode;    while (pahead != pbehind)    {        pahead = pahead->next;        pbehind = pbehind->next;    }    return pahead;}//方法二////判断链表带环情况//pNode CheckCircleEnterNode(pList plist)//{//  pNode fast = plist;//  pNode slow = plist;//  while (fast && fast->next)//  {//      fast = fast->next->next;//      slow = slow->next;//      if (fast == slow)//带环//      {//          return fast;//      }//  }//  return NULL;//}//////求环的长度//int GetCircleLength(pNode meet)//{//  pNode pnode = meet->next;//meet为相遇点//  int count = 1;//  while (pnode != meet)//  {//      pnode = pnode->next;//      count++;//  }//  return count;//}//////求环的入口点//pNode GetCycleEntryNode(pList plist, pNode meet)//{//  pNode pnode = plist;//  if ((plist == NULL) && (meet == NULL))//  {//      return NULL;//  }//  while (pnode != meet)//  {//      meet = meet->next;//      pnode pnode->next;//  }//  return meet;//}void test1(){    pList plist;    pNode ret = NULL;    InitLinkList(&plist);    PushBack(&plist, 1);    PushBack(&plist, 2);    PushBack(&plist, 3);    PushBack(&plist, 4);    PushBack(&plist, 5);    PushBack(&plist, 6);    //ret = FindKTailList(plist, 3);    ret = EnterNodeOfLoop(plist);    Display(plist);    PrintNode(ret);    Destroy(&plist);}int main(){    test1();    system("pause");    return 0;} 

【测试用例】
1、功能测试：
1）链表中包含环或不包含环；
2）链表中有多个或只有一个节点；
2、特殊输入测试：
链表头节点为NULL指针

【心得体会】
要学会把复杂的问题简单化，分布解决问题！

【存在的问题】
1、不会测试带环的链表！

2、//求环的入口点

pNode GetCycleEntryNode(pList plist, pNode meet){    pNode pnode = plist;    if ((plist == NULL) && (meet == NULL))    {        return NULL;    }    while (pnode != meet)    {        meet = meet->next;        pnode pnode->next;    }    return meet;}