Determine if 2 single-linked lists Intersect and find the first intersection node

Without regard to the case of single linked list with ring

If 2 single-linked lists Intersect, it must be a Y-linked list

1. Traverse the 2 linked list to the tail node, record the length of the 2 linked list x,y2. The tail node is the same, then intersect. 3. Starting from the table header, the long list goes first |x-y|, followed by 2 linked lists, to determine the first of the same points.

1#include <stdio.h>2#include <stdlib.h>3#include <string.h>4typedefstructStudent//defining a linked list structure5 {  6     intnum; 7     structStudent *Pnext; 8}stu,*Pstu; 9 voidLink_tail_insert (Pstu, Pstu,int);//establishment of 2 linked lists by tail interpolationTen voidLink_show (PSTU);//Show linked list One voidLink_judge_intersect (Pstu, Pstu);//determine if the linked list has an intersection point A voidLink_bulid_intersect (Pstu, Pstu, Pstu, Pstu *);//set up a Y-linked list - voidMain () { - Pstu Phead1,ptail1,phead2,ptail2;  the     inti;  -Phead1 =NULL;  -Ptail1 =NULL;  -Phead2 =NULL;  +PTAIL2 =NULL;  -      while(SCANF ("%d", &i)! =EOF) {                                        +Link_tail_insert (&phead1,&ptail1,i);  ALink_tail_insert (&phead2,&ptail2,i);  at     }   - link_show (PHEAD1);  -Link_bulid_intersect (phead1,phead2,ptail1,&ptail2);  - link_show (PHEAD2);  - Link_judge_intersect (PHEAD1,PHEAD2);  -System"Pause");  in    - }   to voidLink_tail_insert (Pstu *phead,pstu *ptail,inti) {//The tail interpolation method to establish the linked list + Pstu pnew;  -Pnew = (pstu)malloc(sizeof(Stu));  thememset (Pnew,0,sizeof(Stu));  *Pnew->num =i;  $     if(*ptail = =NULL) {  Panax Notoginseng*phead =pnew;  -*ptail =pnew;  the     }   +     Else{   A(*ptail)->pnext =pnew;  the*ptail =pnew;  +     }   - }   $ voidLink_show (Pstu phead) {//Output Link List $ Pstu pshow;  -Pshow =Phead;  -     if(Phead = =NULL) the     {   -printf"No exsit\n"); Wuyi         return;  the     }   -      while(Pshow! =NULL) {   Wuprintf"%d",pshow->num);  -Pshow = pshow->Pnext;  About     }   $Putchar ('\ n');  - }   - voidLink_bulid_intersect (pstu phead1,pstu phead2,pstu ptail1,pstu *ptail2) {//The 2nd linked list is stitched to the position of the third node of the first list.  - Pstu i;  Ai =Phead1;  +i = i->pnext->Pnext;  the(*PTAIL2)->pnext =i;  -*PTAIL2 =Ptail1;  $ }   the voidLink_judge_intersect (Pstu phead1,pstu phead2) {//determine if the linked list has an intersection point the     intx, y;  the Pstu I,j,prei,prej;  thex = y =0;  -i = Prei = Phead1;//establish the front node, and the current node . inj = Prej =phead2;  the      while(I! = NULL) {//2 linked lists run to tail thePrei =i;  Abouti = i->Pnext;  theX + +;  the     }   the      while(J! =NULL) {   +Prej =J;  -j = j->Pnext;  they++; Bayi     }   the     if(Prei! = Prej)//The tail node is not the same, no intersect. theprintf"Link has no intersect.\n");  -     Else{   -i =Phead1;  thej =phead2;  the         if(x > Y) {//1th Link list length long, first X-y step thex = x-y;  the              while(X >0){   -i = i->Pnext;  thex--;  the             }   the              while(I! = NULL) {//to walk together, to know the encounter or to reach null94                 if(i = =j) {   theprintf"Link has intersect.\n");  theprintf"%d\n",i->num);  the                      Break; 98                 }   Abouti = i->Pnext;  -j = j->Pnext; 101             }  102         }  103         Else{  104y = y-x;//2nd link List length long, first X-y step the              while(Y >0){  106j = j->Pnext; 107y--; 108             }  109              while(I! = NULL) {//to walk together, to know the encounter or to reach null the                 if(i = =j) {  111printf"Link has intersect.\n");  theprintf"%d\n",i->num); 113                      Break;  the                 }   thei = i->Pnext;  thej = j->Pnext; 117             }  118         }  119            -     }  121}

Determine if 2 single-linked lists Intersect and find the first intersection node