[PAT] 02-线性结构2 Reversing Linked List(单向链表的逆转) - C语言实现

　　今天突然想起自己的cnblog有差不多一年没更了😂放一道很久前做的也写好了很久但是一直忘记发布的题.如果有不同的算法欢迎分享~

 

[PAT]02-线性结构2 Reversing Linked List   (25分)

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

Input Specification:

Each input file contains one test case. For each case, the first line contains the address of the first node, a positive NN (\le 10^5≤10​5​​) which is the total number of nodes, and a positive KK (\le N≤N) which is 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 NN 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 is printed in the same format as in the input.

Sample Input:

00100 6 4
00000 4 99999
00100 1 12309
68237 6 -1
33218 3 00000
99999 5 68237
12309 2 33218

Sample Output:

00000 4 33218
33218 3 12309
12309 2 00100
00100 1 99999
99999 5 68237
68237 6 -1

 

 

  1 #include <stdio.h>
  2 #include <malloc.h>
  3 
  4 typedef struct _node
  5 {
  6     int data;
  7     int address;
  8     int back;
  9     int next;
 10 //    int flag;
 11 } NODE;
 12 NODE* List[100001]={0};
 13 int main(void) 
 14 {
 15     int FirstAddr,TotalNode,NumberToReverse,i;
 16     int AddrTemp,DataTemp,NextAddrTemp;
 17     int Loop,Mod;
 18     int count=0;
 19     int Boundary;
 20     NODE* LinkFix=NULL;
 21     NODE* EachNode=NULL;
 22     scanf("%d %d %d",&FirstAddr,&TotalNode,&NumberToReverse);
 23     
 24     int* AddrInOrder=(int *)calloc(TotalNode+1,sizeof(int));
 25     int* LinkReverse=(int *)calloc(TotalNode+1,sizeof(int)); 
 26     for (i=0; i<TotalNode; i++)
 27     {
 28         scanf("%d %d %d",&AddrTemp,&DataTemp,&NextAddrTemp);  
 29         EachNode=(NODE*)calloc(1,sizeof(NODE));
 30         *(List+AddrTemp) = EachNode;
 31         EachNode -> address = AddrTemp;
 32         EachNode -> data = DataTemp;
 33         EachNode -> next = NextAddrTemp;
 34     }
 35 
 36     i=0;
 37     LinkFix = *(List+FirstAddr);
 38     while (1) 
 39     {
 40         *(AddrInOrder+i) = LinkFix -> address;
 41         if (LinkFix -> next == -1) break;
 42     //    (*(List+(LinkFix->next)))->back = LinkFix -> address;
 43         LinkFix = *(List+(LinkFix -> next));
 44         i++;
 45     }
 46     TotalNode=i+1;
 47 //    printf("\n");
 48 //    for (i=0; i<TotalNode; i++) printf("%05d\n",*(AddrInOrder+i));  
 49     Mod = TotalNode % NumberToReverse;      
 50 //    Multily = TotalNode / NumberToReverse;
 51     Loop=NumberToReverse;            
 52     if (NumberToReverse != 1)
 53     {
 54         for (i=0;i<=TotalNode-Mod-1;)
 55         {
 56             if ( count == Loop ) 
 57             {
 58                 count=0;
 59                 i=Loop+i;
 60             }
 61             if (i >= TotalNode-Mod-1) break;
 62             *(LinkReverse+i+count) = *(AddrInOrder+i+(Loop-count-1));
 63             count++;
 64         }
 65         Boundary=i;
 66     }
 67     else  for (i=0;i<=TotalNode-1;i++)    *(LinkReverse+i) = *(AddrInOrder+i);    
 68     
 69     
 70     
 71     if (Mod)
 72     for (i=Boundary; i<TotalNode; i++)
 73     {
 74         *(LinkReverse+i)=*(AddrInOrder+i);
 75         if ( i == TotalNode-1 ) (*(List+*(LinkReverse+i)))->next = -1;
 76     
 77     }
 78     
 79     else
 80     {
 81     //    for (i= TotalNode/2 -1 ; i<TotalNode; i++)    
 82         if (TotalNode == 1) ;
 83         else 
 84         {
 85         (*(List+*(LinkReverse+TotalNode-Loop)))->next = (*(List+*(LinkReverse+TotalNode-Loop+1)))->address;
 86         (*(List+*(LinkReverse+TotalNode-1)))->next = -1;        
 87         (*(List+*(LinkReverse)))->next = (*(List+*(LinkReverse+1)))->address;
 88         }
 89     }
 90     
 91 //    printf("\n");
 92 //    for (k=0; k<TotalNode; k++) printf("%05d\n",*(LinkReverse+k));    
 93     
 94     if (TotalNode>1)
 95     for (i=0; i<TotalNode; i++)
 96     {
 97         if ((*(List+*(LinkReverse+i)))->next == -1) 
 98         {
 99             printf("%.5d %d -1\n",*(LinkReverse+i),(*(List+*(LinkReverse+i)))->data);
100         }
101         else 
102         {
103             printf("%.5d %d %.5d\n",*(LinkReverse+i),(*(List+*(LinkReverse+i)))->data,(*(List+*(LinkReverse+i+1)))->address);
104         }
105         
106     }
107     else printf("%.5d %d -1\n",*(LinkReverse),(*(List+*(LinkReverse)))->data);
108     
109 
110 //    free(AddrInOrder);
111 //    free(LinkReverse);
112     return 0; 
113     
114 }

 

 

提交结果:

测试点6空间消耗很大.以后有时间再优化吧~(立了一个大flag)

posted @ 2017-02-01 15:36  In_Vincible  阅读(1640)  评论(0)  编辑  收藏  举报