判断两个链表是否相交

方法：获得两个链表的长度，获得长度的差值len，然后首先遍历较长的链表len次，然后再同时遍历两个链表，如果有相同部分，两个链表就相交，如果没有，则不相交，即没有公共部分。

代码：

#include <iostream>
#include <cstdlib>
using namespace std;

typedef struct list{
     int data;
     struct list *next;
}*listNode;
void initList(listNode &list)                     //初始化程序
{
     list=(struct list*)malloc(sizeof(struct list));
     list->data=0;
     list->next=NULL;
}

void createList(listNode &list,int a[],int n)        //创建链表
{
    listNode head,node;
    head=list;
    for(int i=0;i<n;i++)
      {
         node=(struct list*)malloc(sizeof(struct list));
         node->data=a[i];
         node->next=NULL;
         head->next=node;
         head=node;
      }
    
}
void judgeInter(listNode list,listNode list1)           //判断两个链表是否相交
{
    int len1=0,len2=0;
    listNode head,head1;
    head=list->next;
    //获得两个链表的长度
    while(head)
    {
       len1++;
       head=head->next;
    }
    head=list1->next;
    while(head)
    {
      len2++;
      head=head->next;
    }
    int t=len1-len2;
    head=list;
    head1=list1;
    while(t>0)
    {
       head=head->next;
       t--;
    }
    while(len2>0)    //判断有没有相同的部分
    {
      head=head->next;
      head1=head1->next;
      if(head->data==head1->data)
       {
           cout<<"the two list have the same part:"<<" "<<head->data<<endl;
           break;
       }
      len2--;
    }
    if(len2<=0)
       cout<<"the two list dont have the same list"<<endl;
}
void playList(listNode list)      //输出单链表
{
    listNode head;
    head=list->next;
    if(head==NULL)
       cout<<"the list is null"<<endl;
    while(head)
     {
       cout<<head->data<<"--";
       head=head->next;
     }
   cout<<endl; 
}

int main()
{
    listNode list,list1;
    int a[10]={1,2,3,4,5,6,7,8,9,0};
    int b[8]={12,13,16,17,11,8,9,0};
    initList(list);
    initList(list1);
    cout<<"create the list:"<<endl;
    createList(list,a,10);
    createList(list1,b,8);
    cout<<"output the list:"<<endl;
    playList(list);
    playList(list1);
    cout<<endl;
    judgeInter(list,list1);
    cout<<endl;
    return 0;
}

运行截图：