算法（3）—— 链表习题 完结

注： 链表有头结点，教科书上的答案会忽略头结点的存在，通过 PrintList  ( List L）就能发现这样的算法存在的问题，解决这个问题需要做两点：（1） 创建节点时候，用 MakeEmpty ( List L ) 创建带头结点的链表； （2） 对链表进行操作的时候，从有数据部分进行操作，即 函数调用参数 为 ( List -> Next)。

1. 给定两个已经排序好的链表 P 和 L ，求二者的交集。 （ 注意，二者未必长度相等 ）

#include <stdio.h>
#include "list.c"


List ListJiaoJi (List P, List L)  {
  if  (P == NULL || L == NULL )
    Error ("Out of space");
  Position   Ppos =  First (P);
  Position   Lpos =  First (L);
  
  List R = MakeEmpty (NULL);
  Position Rpos = Header(R);
 
  while ( Ppos != NULL  && Lpos != NULL )  {

    if  ( Retrieve(Ppos) < Retrieve(Lpos) ) 
      Ppos = Advance (Ppos);
    else
      if ( Retrieve(Ppos) > Retrieve (Lpos)) 
    Lpos = Advance (Lpos);
      else {
    Insert ( Retrieve (Lpos),R,Rpos);
    Lpos = Advance (Lpos);
    Ppos = Advance (Ppos);
    Rpos = Advance (Rpos);
      }
  }
  return R;
}

int main(int argc, char *argv[])   {

  
  List L;
  Position Lpos;
  List P;
  Position Ppos;
  
  int i;
  P = MakeEmpty (NULL);
  L = MakeEmpty( NULL );

  Ppos = Header( P );
  Lpos = Header (L);

  for( i = 0; i < 10; i++ ) {
    Insert( i, L, Lpos );
  
    Lpos = Advance( Lpos );
  }
  PrintList( L );
  
  Insert (3,P,Ppos);
  Ppos = Advance (Ppos);

  Insert (5,P,Ppos);

  PrintList (P);
  List R = MakeEmpty(NULL);
  R =  ListJiaoJi (P,L);
  PrintList (R);
  
  return 0;
}

2.给定两个已经排序好的链表  P 和 L， 求两者的并集。 （ 注意，二者长度未必相等 ）

#include <stdio.h>
#include "list.c"


List ListJiaoJi (List P, List L)  {
  if  (P == NULL || L == NULL )
        Error ("Out of space");
  Position   Ppos =  First (P);
  Position   Lpos =  First (L);
  
  List R = MakeEmpty (NULL);
  Position Rpos = Header(R);
  //PrintList (R);

  while ( Ppos != NULL  && Lpos != NULL )  {

    if  ( Retrieve(Ppos) < Retrieve(Lpos) )  {
      Insert(Retrieve(Ppos),R,Rpos);
      Ppos = Advance (Ppos);
      Rpos = Advance (Rpos);
    }
  
    else if ( Retrieve(Ppos) > Retrieve (Lpos))  {
      Insert ( Retrieve (Lpos),R,Rpos);
      Lpos = Advance (Lpos) ;
      Rpos = Advance (Rpos);
    }

    else {
      Insert ( Retrieve (Lpos),R,Rpos);
      Lpos = Advance (Lpos);
      Ppos = Advance (Ppos);
      Rpos = Advance (Rpos);
    }
  }
    
  while ( Lpos != NULL)  {  
    Insert (Retrieve(Lpos),R,Rpos);
    Lpos = Advance  (Lpos);
    Rpos = Advance (Rpos);
  }

  while ( Ppos != NULL)  {      
    Insert  (Retrieve (Ppos),R,Rpos);
    Ppos = Advance (Ppos);
    Rpos = Advance (Rpos);
  }

 
  return R;
}

int main(int argc, char *argv[])   {

  
  List L;
  Position Lpos;
  List P;
  Position Ppos;
  
  int i;
  P = MakeEmpty (NULL);
  L = MakeEmpty( NULL );

  Ppos = Header( P );
  Lpos = Header (L);

  for( i = 0; i < 10; i++ ) {
    Insert( i, L, Lpos );
  
    Lpos = Advance( Lpos );
  }
  PrintList( L );
  
  Insert (3,P,Ppos);
  Ppos = Advance (Ppos);

  Insert (5,P,Ppos);

  PrintList (P);
  List R = MakeEmpty(NULL);
  R =  ListJiaoJi (P,L);
  PrintList (R);
  
  return 0;
}

3.不用递归方法，反转链表。

树上的答案翻转后的链表并没有头结点，我给改进了一下。

#include "list.c"


/* 

   NOTE: book's answer will just return cur,if do this, the reverselist will not have a head node ,PrintList will not work as hoped.I think it is wrong.It is my method's (as follow) output:
   1 2
   2 1
   1 2 
*/
List ReverseList (List L)  {

  Position pre,cur,nex;

  pre = NULL;
  cur = L->Next;
  nex = First(L)->Next;
  while (nex != NULL )  {
    cur-> Next = pre;
    pre = cur;
    cur = nex;
    nex = Advance(nex);
  }
  cur->Next = pre;       //return cur; 
  nex = MakeEmpty (NULL);
  nex -> Next = cur;

  return nex;
}



int main()  {
  List L = MakeEmpty (NULL);
  Position Pos = Header( L);

  Insert (1,L,Pos);
  Pos = Advance (Pos);

  Insert (2,L,Pos);

  Pos = Advance (Pos);

  PrintList(L);

  List RL =  MakeEmpty(NULL);
  RL = ReverseList (L);
  PrintList (RL);

  List RRL = ReverseList(RL);

  PrintList (RRL);

  return 0;
}

4. 使用递归实现链表的翻转。

#include "list.c"

List  ReverseList (Position P)  {

  if ( P->Next == NULL )  {
     List LR = MakeEmpty (NULL);
     LR -> Next = P;
     return LR;
  }

  List RL = MakeEmpty(NULL);
  RL = ReverseList (P->Next);
  P->Next->Next = P;
  P->Next = NULL;
  return RL;
}    

int main()  {
  List L = MakeEmpty (NULL);
  Position Pos = Header( L);

  Insert (1,L,Pos);
  Pos = Advance (Pos);

  Insert (2,L,Pos);

  Pos = Advance (Pos);

  PrintList(L);

  List RL =  MakeEmpty(NULL);
  RL = ReverseList (L->Next);
  PrintList (RL);

 

  return 0;
}