《C语言实现链表相交、带环问题》

例题：

//1.判断单链表是否带环？若带环，求环的长度？求环的入口点？//2.判断两个链表是否相交，若相交，求交点。（假设链表不带环）//3.判断两个链表是否相交，若相交，求交点。（假设链表可能带环）【升级版】//4.复杂链表的复制。一个链表的每个节点，有一个指向next指针指向下一个节点，//还有一个random指针指向这个链表中的一个随机节点或者NULL，//现在要求实现复制这个链表，返回复制后的新链表。////ps: 复杂链表的结构 //struct ComplexNode//{//  int _data; // 数据 //  struct ComplexNode * _next; // 指向下一个节点的指针 //  struct ComplexNode * _random; // 指向随机节点（可以是链表中的任意节点 or 空） //};

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基本函数：

#include<stdio.h>#include<stdlib.h>#include<assert.h>#include<math.h>typedef int DataType;//节点中数据类型typedef struct ListNode//节点数据结构{    DataType data;    struct ListNode *next;} ListNode;ListNode *Find(ListNode *plist,DataType x)//查找函数，返回查找到节点的地址{    ListNode *cur = plist;    while (cur)    {        if ((cur->data) == x)        {            return cur;        }        cur = cur->next;    }    return cur;}ListNode *BuyNode(DataType x)//产生一个节点并把节点中的数据置数，返回节点地址{    ListNode *plist = (ListNode *)malloc(sizeof(ListNode));    if (plist != NULL)    {        plist->data = x;        plist->next = NULL;        return plist;    }    return NULL;}void PrintList(ListNode *plist)//打印单链表{    assert(plist != NULL);    while (plist)    {        printf("%d->",plist->data);        plist = plist->next;    }    printf("NULL\n");}void PushBuck(ListNode **pplist,DataType x)//在函数尾部插入节点{    ListNode *cur = NULL;    assert(pplist);    cur = *pplist;    if (*pplist == NULL)    {        *pplist = BuyNode(x);        return;    }    while ((cur->next) != NULL)    {        cur = cur->next;    }    cur->next = BuyNode(x);}

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1.判断单链表是否带环？若带环，求环的长度？求环的入口点？

思路：

ListNode *IsCycle(ListNode *plist)//判断是否带环{    ListNode *fast = plist;    ListNode *slow = plist;    assert(plist);    while ((fast != NULL) && (fast->next != NULL))    {        fast = fast->next->next;        slow = slow->next;        if (fast == slow)            return fast;    }    return NULL;}int GetCycleLen(ListNode *meetnode)//求环的长度{    ListNode *cur = NULL;    int count = 1;    assert(meetnode);    cur = meetnode->next;    while (cur != meetnode)    {        cur = cur->next;        count++;    }    return count;}ListNode *GetCycleEnter(ListNode *plist, ListNode *meetnode)//求环的入口点{    assert(plist);    assert(meetnode);    while (plist != meetnode)    {        plist = plist->next;        meetnode = meetnode->next;    }    return plist;}void Test(){    ListNode *plist = NULL;    ListNode *tmp1 = NULL;    ListNode *tmp2 = NULL;    ListNode *tmp = NULL;    PushBuck(&plist, 1);    PushBuck(&plist, 2);    PushBuck(&plist, 3);    PushBuck(&plist, 4);    PushBuck(&plist, 5);    PushBuck(&plist, 6);    PushBuck(&plist, 7);    tmp1 = Find(plist, 3);    tmp2 = Find(plist, 7);    tmp2->next = tmp1;    tmp = IsCycle(plist);    if (tmp == NULL)        printf("不带环！\n");    else    {        printf("带环！\n");        printf("环的长度为：%d\n",GetCycleLen(tmp));        printf("环的入口点为：%d\n",GetCycleEnter(plist,tmp)->data);    }}

结果：

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2.判断两个链表是否相交，若相交，求交点。（假设链表不带环）

bool CheckIsMeet(ListNode *plist1,ListNode *plist2)//判断是否相交{    assert(plist1);    assert(plist2);    while (plist1->next != NULL)    {        plist1 = plist1->next;    }    while (plist2->next != NULL)    {        plist2 = plist2->next;    }    if (plist1 == plist2)    {        return true;    }    else    {        return false;    }}ListNode *AnswerMeetNode(ListNode *plist1,ListNode *plist2)//若相交，求交点{    int count1 = 0;    int count2 = 0;    int count = 0;    ListNode *cur1 = plist1;    ListNode *cur2 = plist2;    while (cur1 != NULL)    {        count1++;        cur1 = cur1->next;    }    while (cur2 != NULL)    {        count2++;        cur2 = cur2->next;    }    count = count1 - count2;    if (count > 0)    {        while (count--)        {            plist1 = plist1->next;        }    }    else    {        count = -count;        while (count--)        {            plist2 = plist2->next;        }    }    while (plist1 != plist2)    {        plist1 = plist1->next;        plist2 = plist2->next;    }    return plist1;}void Test1(){    ListNode *plist1 = NULL;    ListNode *plist2 = NULL;    ListNode *tmp1 = NULL;    ListNode *tmp2 = NULL;    bool flag = true;    PushBuck(&plist1, 1);    PushBuck(&plist1, 3);    PushBuck(&plist1, 5);    PushBuck(&plist1, 7);    PushBuck(&plist1, 8);    PushBuck(&plist1, 9);    PushBuck(&plist1, 10);    PushBuck(&plist2, 2);    PushBuck(&plist2, 4);    PushBuck(&plist2, 6);    tmp1 = Find(plist2, 6);    tmp2 = Find(plist1, 7);    tmp1->next = tmp2;    flag = CheckIsMeet(plist1, plist2);    if (flag)    {        printf("相交！\n");        printf("交点为：%d\n",AnswerMeetNode(plist1,plist2)->data);    }    else    {        printf("不想交！\n");    }}

结果：

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3.判断两个链表是否相交，若相交，求交点。（假设链表可能带环）【升级版】

思路：

RetuenNoode CheckIsMeetTop(ListNode *plist1, ListNode *plist2)//判断两个链表是否相交，若相交，求交点.（假设链表可能带环）【升级版】{    ListNode *tmp1 = NULL;    ListNode *tmp2 = NULL;    RetuenNoode ret = { 0, NULL, NULL };    assert(plist1);    assert(plist2);    tmp1 = IsCycle(plist1);    tmp2 = IsCycle(plist2);    if ((tmp1 == NULL) && (tmp2 == NULL))    {        //都不带环        tmp1 = plist1;        tmp2 = plist2;        while (tmp1->next != NULL)        {            tmp1 = tmp1->next;        }        while (tmp2->next != NULL)        {            tmp2 = tmp2->next;        }        if (tmp1 == tmp2)        {            //相交            ret.plist1 = AnswerMeetNode(plist1, plist2);            ret.flag = 1;            return ret;        }        else        {            //不相交            ret.flag = 1;            return ret;        }    }    else if ((tmp1 != NULL) && (tmp2 != NULL))    {        //两者都带环        ListNode *tmp = tmp1->next;        while ((tmp != tmp2) && (tmp != tmp1))        {            tmp = tmp->next;        }        if (tmp == tmp2)        {            //相交            tmp1 = GetCycleEnter(plist1,tmp1);            tmp2 = GetCycleEnter(plist2,tmp2);            if (tmp1 == tmp2)            {                //交点在链上                int count1 = 0;                int count2 = 0;                int count = 0;                ListNode *cur1 = plist1;                ListNode *cur2 = plist2;                while (cur1 != tmp1)                {                    count1++;                    cur1 = cur1->next;                }                while (cur2 != tmp1)                {                    count2++;                    cur2 = cur2->next;                }                count = count1 - count2;                if (count > 0)                {                    while (count--)                    {                        plist1 = plist1->next;                    }                }                else                {                    count = -count;                    while (count--)                    {                        plist2 = plist2->next;                    }                }                while (plist1 != plist2)                {                    plist1 = plist1->next;                    plist2 = plist2->next;                }                ret.plist1 = plist1;                ret.flag = 3;                return ret;            }            else            {                //两个交点都在环上                ret.plist1 = tmp1;                ret.plist2 = tmp2;                ret.flag = 3;                return ret;            }        }        else        {            //不相交            ret.flag = 3;            return ret;        }    }    else    {        //一个带环，一个不带环。必不相交        ret.flag = 2;        return ret;    }}void Test2(){    RetuenNoode ret = {0,NULL,NULL};    ListNode *plist1 = NULL;    ListNode *plist2 = NULL;    ListNode *tmp1 = NULL;    ListNode *tmp2 = NULL;    PushBuck(&plist1, 1);    PushBuck(&plist1, 3);    PushBuck(&plist1, 5);    PushBuck(&plist1, 7);    PushBuck(&plist1, 8);    PushBuck(&plist1, 9);    PushBuck(&plist1, 10);    PushBuck(&plist2, 0);    PushBuck(&plist2, 2);    PushBuck(&plist2, 4);    PushBuck(&plist2, 6);    tmp1 = Find(plist1, 10);//第三类相交 b> 2>    tmp2 = Find(plist1, 7);    tmp1->next = tmp2;    tmp1 = Find(plist1, 9);    tmp2 = Find(plist2, 6);    tmp2->next = tmp1;    //tmp1 = Find(plist1, 10);//第三类相交 b> 1>    //tmp2 = Find(plist1, 7);    //tmp1->next = tmp2;    //tmp1 = Find(plist1, 5);    //tmp2 = Find(plist2, 6);    //tmp2->next = tmp1;    //tmp1 = Find(plist2, 2);//第三类不相交    //tmp2 = Find(plist2, 6);    //tmp2->next = tmp1;    //tmp1 = Find(plist1, 7);    //tmp2 = Find(plist1, 10);    //tmp2->next = tmp1;    //tmp1 = Find(plist1, 7);//第二类不相交    //tmp2 = Find(plist1, 10);    //tmp2->next = tmp1;    //第一类不相交    //tmp1 = Find(plist2, 6);//第一类相交    //tmp2 = Find(plist1, 7);    //tmp1->next = tmp2;    ret = CheckIsMeetTop(plist1,plist2);    if (ret.flag == 3)    {        if ((ret.plist1) && (ret.plist2))        {            printf(" 第三类 -> b> -> 2> :\nMeetNode1:%d\nMeetNode2:%d\n", ret.plist1->data, ret.plist2->data);        }        else if (ret.plist1)        {            printf("第三类 -> b> -> 1> :\nMeetNode:%d\n", ret.plist1->data);        }        else        {            printf("第三类不相交!\n");        }    }    else if (ret.flag == 1)    {        if (ret.plist1)        {            printf("第一类相交:\nMeetNode:%d\n", ret.plist1->data);        }        else        {            printf("第一类不相交!\n");        }    }    else    {        printf("第二类不相交!\n");    }}int main(){    //Test();    //Test1();    Test2();    return 0;}

结果：

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4.复杂链表的复制。

思路：

#include<stdio.h>#include<stdlib.h>#include<assert.h>typedef int DataType;typedef struct ComplexNode{    DataType data;//数据    struct ComplexNode *next;//指向下一个节点    struct ComplexNode *random;//指向随机节点} ComplexNode;ComplexNode *BuyNode(DataType x)//产生新节点{    ComplexNode *tmp = (ComplexNode *)malloc(sizeof(ComplexNode));    tmp->data = x;    tmp->next = NULL;    tmp->random = NULL;    return tmp;}void PushBack(ComplexNode **pplist,DataType x)//尾插函数{    ComplexNode *cur = NULL;    assert(pplist);    cur = *pplist;    if (*pplist == NULL)    {        *pplist = BuyNode(x);    }    else    {        while (cur->next)        {            cur = cur->next;        }        cur->next = BuyNode(x);    }}ComplexNode *FindComplexNode(ComplexNode *plist,DataType x)//查找函数{    if (plist == NULL)    {        return NULL;    }    else    {        while (plist)        {            if (plist->data == x)                return plist;            plist = plist->next;        }        return NULL;    }}ComplexNode *LineComplexNode(ComplexNode *plist)//第一步、产生节点并连接节点{    ComplexNode *phead = plist;    ComplexNode *tmp = NULL;    ComplexNode *ret = NULL;    assert(plist);      while (plist)    {        ret = plist->next;        tmp = BuyNode(plist->data);        tmp->next = plist->next;        plist->next = tmp;        tmp->random = NULL;        plist = ret;    }    return phead;}void RandomEvaluate(ComplexNode *plist)//第二步、random指针赋值{    assert(plist);    while (plist)    {        if (plist->random == NULL)        {            plist->next->random = NULL;        }        else        {            plist->next->random = plist->random->next;        }        plist = plist->next->next;    }}void PrintComplexNode(ComplexNode *plist)//打印节点{    while (plist)    {        printf("%d->",plist->data);        plist = plist->next;    }    printf("NULL\n");}void PrintRandom(ComplexNode *plist)//打印random指向的数据{    while (plist)    {        if (plist->random == NULL)        {            printf("NULL->");        }        else        {            printf("%d->", plist->random->data);        }        plist = plist->next;    }    printf("STOP\n");}ComplexNode *ToPickComplexList(ComplexNode *plist)//摘出复杂链表{    ComplexNode *phead = NULL;    ComplexNode *cur = NULL;    ComplexNode *tmp = NULL;    if (plist == NULL)    {        return NULL;    }    while (plist)    {        cur = plist->next;        plist->next = cur->next;        if (phead == NULL)            phead = tmp = cur;        else        {            tmp->next = cur;            tmp = cur;        }        plist = plist->next;    }    return phead;}void Test(){    ComplexNode *plist = NULL;    ComplexNode *tmp1 = NULL;    ComplexNode *tmp2 = NULL;    PushBack(&plist, 1);    PushBack(&plist, 2);    PushBack(&plist, 3);    PushBack(&plist, 4);    PushBack(&plist, 5);    tmp1 = FindComplexNode(plist, 3);    plist->random = tmp1;    tmp1->random = NULL;    tmp1 = FindComplexNode(plist, 2);    tmp1->random = plist;    tmp1 = FindComplexNode(plist, 4);    tmp2 = FindComplexNode(plist, 5);    tmp1->random = tmp2;    tmp2->random = tmp2;//初始化复杂链表    LineComplexNode(plist);//第一步    RandomEvaluate(plist);//第二步    tmp1 = ToPickComplexList(plist);//摘出链表    PrintComplexNode(plist);//打印原始链表    PrintRandom(plist);//打印原始Random所指向数据    PrintComplexNode(tmp1);//打印新链表    PrintRandom(tmp1);//打印新random所指向数据}int main(){    Test();    return 0;}

结果：