两个链表的第一个公共节点

题目：输入两个链表，找出它们的第一个公共节点。链表节点定义如下：

typedef int DataType;typedef struct Node{    DataType data;    struct Node *next;}Node, *pList, *pNode;

【方案】由于单向链表的节点只有一个next，所以从第一个节点开始，之后的所有节点都是重合不分叉的，如下图所示。

方案一：在第一条链表上遍历每个节点，每遍历到一个节点就到第二条链表上遍历整条链表，如果两条链表在一个节点上重合，这点即为公共节点，时间复杂度为O(mn)。

方案二：先把两条链表放在两个栈中，这样两条链表的尾节点就位于两个栈的顶部，然后比较两个栈顶的节点。如果相同，则把栈顶弹出再比较下一个栈顶，直到找到最后一个节点。时间复杂度和空间复杂度都是O(m+n),与方案一相比时间效率提高了，但是消耗了一定的空间。

方案三： 第一次遍历两条链表得到它们的长度，及算出长度差；第二次先在长链表上走长度差步，再同时遍历，第一个结合点即为所求。与方案二相比，时间复杂度都是O(m+n)，但是不需要辅助栈，提高了空间效率。

【解题思路】
1、求两条链表的长度及链表长度差。
2、在长链表上走“链表长度差”步。
3、同时遍历长短两条链表。
4、得到第一个公共节点

【代码】

unsigned int GetListLength(pList phead){    unsigned int length = 0;    pNode pnode = phead;    while (pnode != NULL)    {        ++length;        pnode = pnode->next;    }    return length;}pNode ListFirstCommomNode(pList phead1, pList phead2){    //求两条链表的长度及链表长度差    unsigned int length1 = 0;    length1 = GetListLength(phead1);    unsigned int length2 = 0;    length2 = GetListLength(phead2);    int lengthdif = length1 - length2;    pList plistlong = phead1;    pList plistshort = phead2;    if (length1 < length2)    {        plistlong = phead2;        plistshort = phead1;        lengthdif = length2 - length1;      }    //先在长链表上走“链表长度差”步    for (int i = 0; i < lengthdif; ++i)    {        plistlong = plistlong->next;    }    //同时遍历长短两条链表    while ((plistlong != NULL) && (plistshort != NULL) && (plistlong != plistshort))    {        plistlong = plistlong->next;        plistshort = plistshort->next;    }    //得到第一个公共节点//Node pfirstcommomnode = plistlong;    return plistlong;}

【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;}unsigned int GetListLength(pList phead){    unsigned int length = 0;    pNode pnode = phead;    while (pnode != NULL)    {        ++length;        pnode = pnode->next;    }    return length;}pNode ListFirstCommomNode(pList phead1, pList phead2){    //求两条链表的长度及链表长度差步    unsigned int length1 = 0;    length1 = GetListLength(phead1);    unsigned int length2 = 0;    length2 = GetListLength(phead2);    int lengthdif = length1 - length2;    pList plistlong = phead1;    pList plistshort = phead2;    if (length1 < length2)    {        plistlong = phead2;        plistshort = phead1;        lengthdif = length2 - length1;      }    //先在长链表上走“链表长度差”步    for (int i = 0; i < lengthdif; ++i)    {        plistlong = plistlong->next;    }    //同时遍历长短链表    while ((plistlong != NULL) && (plistshort != NULL) && (plistlong != plistshort))    {        plistlong = plistlong->next;        plistshort = plistshort->next;    }    //得到第一个公共节点//Node pfirstcommomnode = plistlong;    return plistlong;}void test1(){    pList plist;    pList plist1;    pList plist2;    pNode ret = NULL;    InitLinkList(&plist);    InitLinkList(&plist1);    InitLinkList(&plist2);    PushBack(&plist2, 6);    PushBack(&plist2, 7);    PushBack(&plist, 1);    PushBack(&plist, 2);    PushBack(&plist, 3);    pList tmp = plist;    while (tmp->next)    {        tmp = tmp->next;    }    tmp->next = plist2;    Display(plist);    PushBack(&plist1, 4);    PushBack(&plist1, 5);    pList cur = plist1;    while (cur->next)    {        cur = cur->next;    }    cur->next = plist2;    Display(plist1);    ret = ListFirstCommomNode(plist, plist1);    PrintNode(ret);    Destroy(&plist);}int main(){    test1();    system("pause");    return 0;}

运行结果：

【测试】
1、功能测试
1）输入的两条链表没有公共节点；
2）第一个公共节点在链表的中间；
3）第一个公共节点在链表的头节点；
4）第一个公共节点在链表的尾节点；

2、特殊输入测试
输入的链表头节点是NULL指针