PAT1074. Reversing Linked List

Given a constant k and a singly linked list L, you is supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K = 3, then you must output 3→2→1→6→5→4; If K = 4, you must output 4→3→2→1→5→6.

Input Specification:

Each input file contains the one test case. The first line contains the address of the first node, a positive N (<=) which are the total number O  F nodes, and a positive K (<=n) which are the length of the sublist to be reversed. The address of a node is a 5-digit nonnegative integer, and NULL is represented by-1.

Then N lines follow, each describes a node in the format:

Address Data Next

Where Address is the position of the node, Data is an integer, and Next is the position of the Next node.

Output Specification:

For each case, output the resulting ordered linked list. Each node occupies a line, and was printed in the same format as in the input.

Sample Input:

00100 6 400000 4 9999900100 1 1230968237 6-133218 3 0000099999 5 6823712309 2 33218

Sample Output:

00000 4 3321833218 3 1230912309 2 0010000100 1 9999999999 5 6823768237 6-1
Thinking: This problem I spent a long time to parse, why spent a long time, first did not read the problem, and then the foundation of the lack of reading and the struggle for a long time is the temporary variable. The final reading of the question on the basis of no consideration of the boundary situation. In short: Benefit


1: This is the wrong code: the wrong code must be remembered for a lifetime.

1#include <iostream>2#include <cstdio>3 using namespacestd;4 #defineMAX 1000105 structInfo6 {7     intPre; 8     intdata; 9     intNext;Ten }node[max]; One //Be sure to read the question carefully .... every k elements A intMainintargcChar*argv[]) - { -     intstart,n,k; thescanf"%d%d%d",&start,&n,&K); -      for(intI=0; i<n;i++) -     { -         intPre,data,next; +scanf"%d%d%d",&pre,&data,&next); -Node[pre].data=data; +node[pre].next=Next; A     } at     intpt=start; -     //record the precursor -node[pt].pre=-2;//the beginning of the precursor represents the beginning -      while(node[pt].next!=-1) -     { -         inttemp=pt; inpt=Node[pt].next; -Node[pt].pre=temp; to     }      +     //looking for the first K -pt=start; the      for(intI=1; i<k&&node[pt].next!=-1; i++) *     { $pt=Node[pt].next;Panax Notoginseng     } -     //to reverse the     inttemp=Node[pt].next; +start=pt; A      while(node[pt].pre!=-2) the     { +node[pt].next=Node[pt].pre; -pt=Node[pt].pre; $     }  $node[pt].next=temp; -     //output of the results -pt=start; the      while(node[pt].next!=-1) -     {Wuyiprintf"%05d%d%05d\n", pt,node[pt].data,node[pt].next); thept=Node[pt].next; -     }  Wuprintf"%05d%d%d\n", pt,node[pt].data,node[pt].next); -     return 0;

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2: The correct code

1#include <cstdio>2#include <vector>3 using namespacestd;4 #defineMAX 1000105 structnode6 {7     intadd;8     intdata;9     intNext;Ten }; One intMain () A { -Vector<node>inch(MAX);//like an array -Vector<node>SRT; theVector<node> out; -     intstart,n,k; -scanf"%d%d%d",&start,&n,&K); -      for(intI=0; i<n;i++) +     { - node temp; +scanf"%d%d%d",&temp.add,&temp.data,&temp.next); A         inch[temp.add]=temp; at     } -     intnext=start; -      while(next!=-1) -     { -Srt.push_back (inch[next]); -next=inch[Next].next; in     } -     intleft=0; to      while(left<srt.size ()) +     { -          the          for(inti=left+k-1; i>=left;i--) *         { $             if(i>=srt.size ())Panax Notoginseng                Break; -              out. Push_back (Srt[i]); the         } +left+=K; A     } the     if(left!=srt.size ()) +left-=K; -      while(left<srt.size ()) $     { $          out. Push_back (Srt[left]); -left++; -     } the     /*  - int len=srt.size ();Wuyi int right=k-1; the While (Right<len) -     { Wu for (int i=right;i>right-k;i--) -        { About Out.push_back (Srt[i]);  $        } - right+=k;  -     } - right-=k; A right++; + for (int i=right;i<srt.size (); i++) the     { - Out.push_back (Srt[i]); $     }*/  the      for(intI=0;i< out. Size (); i++) the     { the         if(i== out. Size ()-1) theprintf"%05d%d-1\n", out[I].add, out[I].data]; -         Else inprintf"%05d%d%05d\n", out[I].add, out[I].data, out[i+1].add); the     } the     return 0; About}

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PAT1074. Reversing Linked List