找出链表的第一个公共结点 【微软面试100题 第六十二题】

题目要求：

　　两个无环单向链表，找出它们的第一个公共结点。

　　参考资料：剑指offer第37题。

题目分析：

　　

代码实现：

　　

#include <iostream>

using namespace std;

typedef struct ListNode
{
    struct ListNode *next;
    int data;
}ListNode;

void InitList(ListNode **head1,ListNode **head2);
ListNode *FindFirstCommonNode(ListNode *pHead1,ListNode *pHead2);
void PrintList(ListNode *h);

int main(void)
{
    ListNode *h1 = NULL,*h2 = NULL;
    
    InitList(&h1,&h2);
    cout << "h1为：";
    PrintList(h1);
    cout << "h2为：";
    PrintList(h2);
    ListNode *pFirstCommonNode = FindFirstCommonNode(h1,h2);
    if(pFirstCommonNode)
        cout << "第一个公共结点的值为：" << pFirstCommonNode->data << endl;
    else
        cout <<"没有公共结点" << endl;
    return 0;
}
void PrintList(ListNode *h)
{
    while(h!=NULL)
    {
        cout << h->data << " ";
        h = h->next;
    }
    cout << endl;
}
unsigned int GetListLength(ListNode *pHead)
{
    unsigned int nLen = 0;
    ListNode *pNode = pHead;
    while(pNode != NULL)
    {
        ++nLen;
        pNode = pNode->next;
    }
    return nLen;
}
ListNode *FindFirstCommonNode(ListNode *pHead1,ListNode *pHead2)
{
    if(pHead1==NULL || pHead2==NULL)
        return NULL;
    unsigned int len1 = GetListLength(pHead1);
    unsigned int len2 = GetListLength(pHead2);
    int lenDif = len1-len2;
    ListNode *pListHeadLong = pHead1;
    ListNode *pListHeadShort = pHead2;
    if(len1<len2)
    {
        pListHeadLong = pHead2;
        pListHeadShort = pHead1;
        lenDif = len2-len1;
    }
    for(int i = 0;i<lenDif;i++)
        pListHeadLong = pListHeadLong->next;
    while((pListHeadLong!=NULL) && (pListHeadShort!=NULL) && (pListHeadLong!=pListHeadShort))
    {
        pListHeadLong = pListHeadLong->next;
        pListHeadShort = pListHeadShort->next;
    }
    return pListHeadLong;
}
//head1:1-->5-->9-->NULL
//head2:2-->4-->10-->5-->9-->NULL
void InitList(ListNode **head1,ListNode **head2)
{
    ListNode *tmp = new ListNode,*store;
    tmp->data = 1;
    *head1 = tmp;

    tmp = new ListNode;
    tmp->data = 5;
    store = tmp;
    (*head1)->next = tmp;

    ListNode *tmp1 = new ListNode;
    tmp1->data = 9;
    tmp1->next = NULL;
    tmp->next = tmp1;

    tmp = new ListNode;
    tmp->data = 2;
    *head2 = tmp;

    tmp = new ListNode;
    tmp->data = 4;
    (*head2)->next = tmp;

    tmp1 = new ListNode;
    tmp1->data = 10;
    tmp1->next = store;
    tmp->next = tmp1;
}