算法（2）—— 链表习题

题目:0：给定链表 L 和链表 P ，它们包含以升序排列的整数。操作 PrintList  ( L , P )，将打印 L 中那些由 P 所指定位置上的元素，如 P 中的元素为 1 4 5，则打印 L 中第 1,4,5个元素。

解法1： 计算记录每次 L 中遍历的步长 来打印相应节点的元素。

解法2： 循环遍历 L 并且 由整形变量记录走到第 n 个元素处，若这个 n 与 P的记录的数据相同，则打印  L 中该节点的数据。

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

void PrintPos (List L,List P)  {
  Position Ppos,Lpos;
  Ppos = First(P);
  Lpos = First(L);

  int pre,now,step;
  pre = 1;
  
  while (Lpos != NULL && Ppos !=  NULL)  {

    now = Ppos->Element;

    step = now - pre;
    for (int i = 0; i < step ; i++)
      Lpos = Advance (Lpos);

    printf ("%d  ",Lpos->Element);
    pre = now;
    Ppos = Advance(Ppos);
  }
}



int main()  {

    List L = MakeEmpty(NULL);
    Position Lpos = Header(L);

    List P = MakeEmpty(NULL);
    Position Ppos = Header(P);
  
    for (int i = 0;i < 10;i++)  {
      Insert(i,L,Lpos);
      Lpos = Advance(Lpos);
    }

    Insert(1,P,Ppos);
    Ppos = Advance(Ppos);
    Insert(3,P,Ppos);
    Ppos = Advance(Ppos);
    Insert(9,P,Ppos);
    
    PrintPos(L,P);
    
    return 0;
 }

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

void PrintPos (List L, List P)  {
  Position Lpos = Header (L);
  Position Ppos = First (P);

  int vstl=0;

  while ( Lpos != NULL && Ppos != NULL )  {
    if  (vstl++ == Ppos->Element)  {

      if (Ppos->Element <= 0)  ;
      else printf ("%d ",Lpos->Element);

      Ppos= Advance(Ppos);
    }

    Lpos = Advance(Lpos);
  }
}
     
int main()  {

    List L = MakeEmpty(NULL);
    Position Lpos = Header(L);

    List P = MakeEmpty(NULL);
    Position Ppos = Header(P);
  
    for (int i = 0;i < 10;i++)  {
      Insert(i,L,Lpos);
      Lpos = Advance(Lpos);
    }

    Insert(1,P,Ppos);
    Ppos = Advance(Ppos);
    Insert(3,P,Ppos);
    Ppos = Advance(Ppos);
    Insert(9,P,Ppos);
    
    PrintPos(L,P);
    
    return 0;
 }

题目1：通过只调整指针交换单链表的的两个节点

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

void SwapNode (List L, Position P1, Position P2)  {
  Position pre = FindPrevious(P1->Element,L);
  Position tmp = Advance(P2);

  pre-> Next = P2;
  P2 -> Next = P1;
  P1 -> Next = tmp;

}

int main()  {

    List L = MakeEmpty(NULL);
    Position Lpos = Header(L);

     
    for (int i = 0;i < 10;i++)  {
      Insert(i,L,Lpos);
      Lpos = Advance(Lpos);
    }

    PrintList(L);
    printf("
");

    Position P1 = Find (3,L);
    Position P2 = Advance (P1);
    SwapNode (L,P1,P2);

    PrintList(L);
     
    return 0;
 }