56. 链表中环的入口结点

题目：一个链表中包含环，如何找出环的入口结点？例如，在图8.3的链表中，环的入口结点是结点3。

思路：定义两个指针p1,p2都指向头结点，假设链表中有n个结点，先让p1先走n个结点，然后让p1,p2同时走，当两个指针相遇时，相遇的结点就是环的入口结点。

解题步骤：(1)：先判断链表中是否包含环。定义两个指针，一个慢指针每次走1步，一个快指针每次走2步，当两个指针能相等时，也就是相遇，链表中就含有环。(2)：计算出环的结点数。用(1)中相遇的结点再继续走，定义一个变量n每次进行自增，当能碰到它自己的话，n就是环的结点数。(3)：用思路进行计算即可。

#include<iostream>
#include<stack>
#include<cstdio>
using namespace std;

struct ListNode
{
    int                 m_nValue;
    ListNode*           m_pNext;
};

ListNode* CreateListNode(int value)
{
    ListNode* node = new ListNode();
    node->m_nValue = value;
    node->m_pNext = NULL;

    return node;
}

void ConnectListNodes(ListNode* node1, ListNode* node2)
{
    if (node1 != NULL && node2 != NULL)
    {
        node1->m_pNext = node2;
    }
}

void PrintListNode(ListNode* pHead)
{
    if (pHead == NULL)
    {
        return;
    }

    if (pHead != NULL)
    {
        cout << pHead->m_nValue << " ";
        pHead = pHead->m_pNext;
    }

    cout << endl;

}

void DestroyList(ListNode* pHead)
{
    ListNode* node = pHead;

    while (node != NULL)
    {
        pHead = pHead->m_pNext;
        delete node;
        node = pHead;
    }
}





ListNode* MeetingNode(ListNode* pHead)
{
    if (pHead == NULL)
    {
        return NULL;
    }

    ListNode* pSlow = pHead->m_pNext;
    if (pSlow == NULL)
    {
        return NULL;
    }

    ListNode* pFast = pSlow->m_pNext;
    while (pFast != NULL && pSlow != NULL)
    {
        if (pFast == pSlow)
        {
            return pFast;
        }

        pSlow = pSlow->m_pNext;
        pFast = pFast->m_pNext;

        if (pFast != NULL)
        {
            pFast = pFast->m_pNext;
        }

    }

    return NULL;

}

ListNode* EntryNodeOfLoop(ListNode* pHead)
{
    ListNode* meetingNode = MeetingNode(pHead);
    if (meetingNode == NULL)
    {
        return NULL;
    }

    int nodesInLoop = 1;
    ListNode* pNode1 = meetingNode;
    while (pNode1->m_pNext != meetingNode)
    {
        pNode1 = pNode1->m_pNext;
        ++nodesInLoop;
    }

    pNode1 = pHead;
    for (int i = 0; i < nodesInLoop; i++)
    {
        pNode1 = pNode1->m_pNext;
    }

    ListNode* pNode2 = pHead;
    while (pNode1 != pNode2)
    {
        pNode1 = pNode1->m_pNext;
        pNode2 = pNode2->m_pNext;
    }

    return pNode1;

}


void test1()
{
    ListNode* node1 = CreateListNode(1);
    ListNode* node2 = CreateListNode(2);
    ListNode* node3 = CreateListNode(3);
    ListNode* node4 = CreateListNode(4);
    ListNode* node5 = CreateListNode(5);
    ListNode* node6 = CreateListNode(6);
    ListNode* node7 = CreateListNode(7);

    ConnectListNodes(node1, node2);
    ConnectListNodes(node2, node3);
    ConnectListNodes(node3, node4);
    ConnectListNodes(node4, node3);

    ListNode* node = EntryNodeOfLoop(node1);
    PrintListNode(node);

}

int main()
{
    test1();

    return 0;
}