《剑指offer》第二十三题（链表中环的入口结点）

// 面试题23：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3.8的链表中，
// 环的入口结点是结点3。

#include <iostream>
#include "List.h"

ListNode* MeetingNode(ListNode* pHead)//鲁棒一：先确定有没有环，有的话先求得环中任意一个节点
{
    if (pHead == nullptr)//若头结点为空
        return nullptr;

    ListNode* pSlow = pHead->m_pNext;
    if (pSlow == nullptr)//若只有一个节点
        return nullptr;

    ListNode* pFast = pSlow->m_pNext;
    while (pFast != nullptr && pSlow != nullptr)//判断碰到尾节点后有没有环
    {
        if (pFast == pSlow)
            return pFast;

        pSlow = pSlow->m_pNext;

        pFast = pFast->m_pNext;
        if (pFast != nullptr)//判断碰到尾节点后有没有环
            pFast = pFast->m_pNext;
    }

    return nullptr;
}

ListNode* EntryNodeOfLoop(ListNode* pHead)//开始计算入口节点，第一步先求环的个数，第二步通过两个前后指针计算入口节点
{
    ListNode* meetingNode = MeetingNode(pHead);
    if (meetingNode == nullptr)//如果存在环，得到一个环中节点
        return nullptr;

    // 得到环中结点的数目
    int nodesInLoop = 1;
    ListNode* pNode1 = meetingNode;
    while (pNode1->m_pNext != meetingNode)
    {
        pNode1 = pNode1->m_pNext;
        ++nodesInLoop;
    }

    // 先移动pNode1，次数为环中结点的数目
    pNode1 = pHead;
    for (int i = 0; i < nodesInLoop; ++i)
        pNode1 = pNode1->m_pNext;

    // 再移动pNode1和pNode2
    ListNode* pNode2 = pHead;
    while (pNode1 != pNode2)//二者只能相遇在入口处
    {
        pNode1 = pNode1->m_pNext;
        pNode2 = pNode2->m_pNext;
    }

    return pNode1;
}

// ==================== Test Code ====================
void Test(const char* testName, ListNode* pHead, ListNode* entryNode)
{
    if (testName != nullptr)
        printf("%s begins: ", testName);

    if (EntryNodeOfLoop(pHead) == entryNode)
        printf("Passed.\n");
    else
        printf("FAILED.\n");
}

// A list has a node, without a loop
void Test1()
{
    ListNode* pNode1 = CreateListNode(1);

    Test("Test1", pNode1, nullptr);

    DestroyList(pNode1);
}

// A list has a node, with a loop
void Test2()
{
    ListNode* pNode1 = CreateListNode(1);
    ConnectListNodes(pNode1, pNode1);

    Test("Test2", pNode1, pNode1);

    delete pNode1;
    pNode1 = nullptr;
}

// A list has multiple nodes, with a loop 
void Test3()
{
    ListNode* pNode1 = CreateListNode(1);
    ListNode* pNode2 = CreateListNode(2);
    ListNode* pNode3 = CreateListNode(3);
    ListNode* pNode4 = CreateListNode(4);
    ListNode* pNode5 = CreateListNode(5);

    ConnectListNodes(pNode1, pNode2);
    ConnectListNodes(pNode2, pNode3);
    ConnectListNodes(pNode3, pNode4);
    ConnectListNodes(pNode4, pNode5);
    ConnectListNodes(pNode5, pNode3);

    Test("Test3", pNode1, pNode3);

    delete pNode1;
    pNode1 = nullptr;
    delete pNode2;
    pNode2 = nullptr;
    delete pNode3;
    pNode3 = nullptr;
    delete pNode4;
    pNode4 = nullptr;
    delete pNode5;
    pNode5 = nullptr;
}

// A list has multiple nodes, with a loop 
void Test4()
{
    ListNode* pNode1 = CreateListNode(1);
    ListNode* pNode2 = CreateListNode(2);
    ListNode* pNode3 = CreateListNode(3);
    ListNode* pNode4 = CreateListNode(4);
    ListNode* pNode5 = CreateListNode(5);

    ConnectListNodes(pNode1, pNode2);
    ConnectListNodes(pNode2, pNode3);
    ConnectListNodes(pNode3, pNode4);
    ConnectListNodes(pNode4, pNode5);
    ConnectListNodes(pNode5, pNode1);

    Test("Test4", pNode1, pNode1);

    delete pNode1;
    pNode1 = nullptr;
    delete pNode2;
    pNode2 = nullptr;
    delete pNode3;
    pNode3 = nullptr;
    delete pNode4;
    pNode4 = nullptr;
    delete pNode5;
    pNode5 = nullptr;
}

// A list has multiple nodes, with a loop 
void Test5()
{
    ListNode* pNode1 = CreateListNode(1);
    ListNode* pNode2 = CreateListNode(2);
    ListNode* pNode3 = CreateListNode(3);
    ListNode* pNode4 = CreateListNode(4);
    ListNode* pNode5 = CreateListNode(5);

    ConnectListNodes(pNode1, pNode2);
    ConnectListNodes(pNode2, pNode3);
    ConnectListNodes(pNode3, pNode4);
    ConnectListNodes(pNode4, pNode5);
    ConnectListNodes(pNode5, pNode5);

    Test("Test5", pNode1, pNode5);

    delete pNode1;
    pNode1 = nullptr;
    delete pNode2;
    pNode2 = nullptr;
    delete pNode3;
    pNode3 = nullptr;
    delete pNode4;
    pNode4 = nullptr;
    delete pNode5;
    pNode5 = nullptr;
}

// A list has multiple nodes, without a loop 
void Test6()
{
    ListNode* pNode1 = CreateListNode(1);
    ListNode* pNode2 = CreateListNode(2);
    ListNode* pNode3 = CreateListNode(3);
    ListNode* pNode4 = CreateListNode(4);
    ListNode* pNode5 = CreateListNode(5);

    ConnectListNodes(pNode1, pNode2);
    ConnectListNodes(pNode2, pNode3);
    ConnectListNodes(pNode3, pNode4);
    ConnectListNodes(pNode4, pNode5);

    Test("Test6", pNode1, nullptr);

    DestroyList(pNode1);
}

// Empty list
void Test7()
{
    Test("Test7", nullptr, nullptr);
}

int main(int argc, char* argv[])
{
    Test1();
    Test2();
    Test3();
    Test4();
    Test5();
    Test6();
    Test7();
    system("pause");
    return 0;
}