判断2个单链表是否相交，并求出第一个相交结点

不考虑单链表有环的情况下

如果2个单链表相交,一定是Y型链表

1.遍历2个链表到尾结点,记录2个链表的长度x,y

2.尾结点相同，则相交。

3.从表头开始，长链表先走|x-y|步，之后2个链表一起走，判断第一个相同的点。

 

#include <stdio.h>  
#include <stdlib.h>  
#include <string.h>  
typedef struct student             //定义链表结构  
{  
    int num;  
    struct student *pnext;  
}stu,*pstu;  
void link_tail_insert(pstu ,pstu ,int );                        //尾插法建立2个链表  
void link_show(pstu );                                                       //展示链表  
void link_judge_intersect(pstu ,pstu );                          //判断链表是否有交点  
void link_bulid_intersect(pstu ,pstu ,pstu ,pstu *); //建立Y型链表  
void main(){  
    pstu phead1,ptail1,phead2,ptail2;  
    int i;  
    phead1 = NULL;  
    ptail1 = NULL;  
    phead2 = NULL;  
    ptail2 = NULL;  
    while(scanf("%d",&i) != EOF){                                       
        link_tail_insert(&phead1,&ptail1,i);  
        link_tail_insert(&phead2,&ptail2,i);  
    }  
    link_show(phead1);  
    link_bulid_intersect(phead1,phead2,ptail1,&ptail2);  
    link_show(phead2);    
    link_judge_intersect(phead1,phead2);  
    system("pause");  
  
}  
void link_tail_insert(pstu *phead,pstu *ptail,int i){              //尾插法建立链表  
    pstu pnew;  
    pnew = (pstu)malloc(sizeof(stu));  
    memset(pnew,0,sizeof(stu));  
    pnew->num = i;  
    if(*ptail == NULL){  
        *phead = pnew;  
        *ptail = pnew;  
    }  
    else{  
        (*ptail)->pnext = pnew;  
        *ptail = pnew;  
    }  
}  
void link_show(pstu phead){                                         //输出链表  
    pstu pshow;  
    pshow = phead;  
    if(phead == NULL)  
    {  
        printf("no exsit\n");  
        return;  
    }  
    while(pshow != NULL){  
        printf("%d ",pshow->num);  
        pshow = pshow->pnext;  
    }  
    putchar('\n');  
}  
void link_bulid_intersect(pstu phead1,pstu phead2,pstu ptail1,pstu *ptail2){   //把第2个链表拼接到第一个链表第三个结点的位置。  
    pstu i;  
    i = phead1;  
    i = i->pnext->pnext;            
    (*ptail2)->pnext = i;  
    *ptail2 = ptail1;  
}  
void link_judge_intersect(pstu phead1,pstu phead2){             //判断链表是否有交点  
    int x,y;  
    pstu i,j,prei,prej;  
    x = y = 0;  
    i = prei = phead1;                                        //建立前置结点，和当前结点  
    j = prej = phead2;  
    while(i != NULL){                                        //2个链表运行到尾部  
        prei = i;  
        i = i->pnext;  
        x++;  
    }  
    while(j != NULL){  
        prej = j;  
        j = j->pnext;  
        y++;  
    }  
    if(prei != prej)                                      //尾结点不相同 没有相交  
        printf("Link has no intersect.\n");  
    else{  
        i = phead1;  
        j = phead2;  
        if(x > y){                                        //第1个链表长度长，先走X-Y步  
            x = x - y;  
            while(x > 0){  
                i = i->pnext;  
                x--;  
            }  
            while(i != NULL){                              //一起走，知道相遇或者达到NULL  
                if(i == j){  
                    printf("Link has intersect.\n");  
                    printf("%d\n",i->num);  
                    break;  
                }  
                i = i->pnext;  
                j = j->pnext;  
            }  
        }  
        else{  
            y = y - x;                           //第2个链表长度长，先走X-Y步  
            while(y > 0){  
                j = j->pnext;  
                y--;  
            }  
            while(i != NULL){                       //一起走，知道相遇或者达到NULL  
                if(i == j){  
                    printf("Link has intersect.\n");  
                    printf("%d\n",i->num);  
                    break;  
                }  
                i = i->pnext;  
                j = j->pnext;  
            }  
        }  
          
    }  
}