< written >< interview >c/c++ single-linked list (4) determine if the two linked lists intersect, find the intersection (linked list without ring/May band ring)

Determine if the two linked lists intersect, to find the intersection (assuming that the linked list does not have a ring)

Determine if the two linked lists intersect to find the intersection (assuming that the linked list may have loops)

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Ringentry_point () and other functions see the previous chapter.

Slistnode* intersect (SLISTNODE *&L, SLISTNODE *&M)//To determine whether the two linked lists intersect, to find the intersection (assuming the linked list without a ring) { / /idea: If there is no ring, only intersect/do not want to cross two cases  //  the same as Ringentry_point () function method: //      Two list length difference K, Let a pointer start with a long list of K-steps, and another pointer starting from the short-linked list, //      two linked list to go together, the meeting point is the entry point  if  (l !=  null&&m != null)  {  SListNode *cur = L;   Slistnode *newnode = m;  int c1 = _lengthnode (cur);   int  c2 = _lengthnode (NewNode);   int minus = _lengthnode (cur)  -  _lengthnode (NewNode);  if  (minus > 0)   {   while  ( minus--)    {    cur = cur->next;   }   }  else  {   int tmp = -minus;   while   (tmp--)     {    newnode = newnode->next;   }  }   while  (newnode != cur&&newnode != null&&cur != null)    {   NewNode = NewNode->next;   cur = cur->next;   }  if  (newnode == null | |  cur == null)    return null;  return cur; } return  null;}  slistnode* intersectring (SLISTNODE *&L, SLISTNODE *&M)//Determine whether the two linked lists Intersect, Finding the intersection (assuming that the chain list may have a ring) { //thought: Consider the whole case: 1, without ring   intersect/disjoint   2, with ring   one band ring without ring/two band ring (where the intersection is not on the ring/intersection on the ring)    //distinguish between the circumstances, respectively, to achieve the  if  (l != null&&m != null)  {  if   (! IsRing (L)  && ! IsRing (M))//Without ring    return intersect (l, m);  else if  ((IsRing (L)  && ! IsRing (M))  | |   (! IsRing (L)  && isring (M))//one strip without ring    return NULL;  else   {   //first to find the ring entry point of the two linked lists, if the entry point is the same, then the 5th case is 4th, 6 kinds of circumstances    slistnode *re1 =  ringentry (L);    slistnode *re2 = ringentry (M);   if  ( Re1 == re2)//5th case   The problem of cross-linked list without ring    {    SLISTNODE* M1  = isring (L);    slistnode*recover = m1->next;//  restoration modeled after Ringentry _point () function     m1->next = null;    slistnode*cur =  intersect (l, m);    m1->next = recover;     return cur;   }   else//4th, 6 cases if one of the ring entry points is on another ring, it is the 6th case, otherwise the 4th case     {    slistnode *tmp = re1->next;    while&nBSP; (tmp != re1&&tmp != re2)      tmp = tmp-> next;    if  (Tmp == re1)//4th case      return  null;    printf ("Two-chain bracelet, with two intersections \ n");//6th Case     SLISTNODE *TMP1  = re1->next;    re1->next = NULL;     Printnode (Re1);    re1->next = tmp1;    return re2;    }  } } else  return null;}

Test

Void test10 () { printf ("//test10 ()  intersect ()  \n"); slistnode *ll =  Null; pushback (ll, 1);  pushback (ll, 2);  pushback (ll, 3);  PushBack (LL,  4);  pushback (ll, 5);  printnode (LL);  slistnode *mm = null; pushback (MM,  1);  pushback (mm, 2); mm->next->next = ll->next->next->next;  printnode (MM);  slistnode *nn = null; pushback (nn, 7);  PushBack (NN,  8);  pushback (nn, 9);  pushback (nn, 0);  printnode (NN);  slistnode* c1  = intersect (LL, MM);  slistnode* c2 = intersect (LL, NN);  Printnode (C1);  printnode (C2);} Void test11 () { printf ("//test11 ()  intersectring ()  \n");  slistnode *ll =  null; pushback (ll, 1);  pushback (ll, 2);  pushback (ll, 3);  PushBack (LL, 4);  pushback (ll, 5);  printnode (LL); slistnode *nn = null;  Pushback (nn, 6);   pushback (nn, 7);  pushback (nn, 8);  pushback (NN, 9);  pushback (nn, 0);  printnode (NN);  slistnode* c1 = intersectring (LL, NN) ;//1th  printnode (C1),  printf ("\ n"),  slistnode *mm = null;//2nd  pushback (MM,  0);  pushback (mm, 1); mm->next->next = ll->next; slistnode*  C2 = intersectring (LL, MM);  printnode (C2);  printf ("\ n");  find (LL, 5) Next = find (ll, 3);//3rd Type  slistnode* c3 = intersectring (LL, NN);  Slistnode* c4 = intersectring (NN, LL);  printnode (C3);  printnode (C4);  printf ("\ n");  find (nn, 0)->next = find (nn, 8);//4th  SListNode* c5 =  Intersectring (nn,  LL);  printnode (C5);  printf ("\ n");  slistnode* c6 = intersectring (MM, LL);// 5th species of  slistnode* c7 = intersectring (ll, mm);  slistnode* tmp1 = c6- >next; c6->next = null; printnode (C6); c6->next = tmp1;  Slistnode* tmp2 = c7->next; c7->next = null; printnode (C7);  c7- >next = tmp2; printf ("\ n");  find (nn, 0)->next = find (LL, 4);// 6th species of  slistnode* c9 = intersectring (NN, LL);  slistnode* tmp3 = c9- >next; c9->next = null; printnode (C9); c9->next = tmp3;  Slistnode* c10 = intersectring (LL, NN);  slistnode* tmp4 = c10->next ;  c10->next = null; printnode (C10);  c10->next = tmp4; printf ("\ n ");  find (ll, 5)->next = find (ll, 1);//6th  find (nn, 0)->next = find (LL, 4);  slistnode* c11 = intersectring (NN, LL);  slistnode* tmp5 = c11- >next; c11->next = null; printnode (C11); c11->next = tmp5;  Slistnode* c12 = intersectring (LL, NN);  slistnode* tmp6 = c12->next ;  c12->next = null; printnode (C12);  c12->next = tmp6; printf ("\ n ");}

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< written >< interview >c/c++ single-linked list (4) determine if the two linked lists intersect, find the intersection (linked list without ring/May band ring)