PAT 02-线性结构1. Reversing Linked List

时间：2015-07-07 11:02:57 阅读：359 评论：0 收藏：0 [点我收藏+]

标签：

Given a constant K and a singly linked list L, you are 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 one test case. For each case, the first line contains the address of the first node, a positive N (<= 105) which is the total number of nodes, and a positive K (<=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 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 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 #define MAX  100004
  3 
  4 typedef struct List {
  5     int Data;
  6     int Addr;
  7     int NextAddr;
  8     struct List *Next;
  9 }iList;
 10 
 11 void PrintList (iList *a);
 12 iList *ListReversing (iList *p,int K);
 13 
 14 int main()
 15 {
 16     int FirstAddr;
 17     int i;
 18     int K;   //反转子链长度K
 19     int N;  // the total number of nodes
 20     int Num; //链表建好之后的节点数
 21     int data[MAX];
 22     int NextAddress[MAX];
 23     int temp;
 24 
 25 scanf("%d %d %d",&FirstAddr,&N,&K);
 26 
 27 iList a[N+1];
 28 
 29 a[0].NextAddr = FirstAddr;
 30 
 31 for(i = 0; i<N ; i++)
 32     {
 33         scanf("%d",&temp);
 34         scanf("%d %d",&data[temp],&NextAddress[temp]);
 35     }
 36 
 37 i = 1;
 38     while(1)
 39     {
 40         if(a[i-1].NextAddr == -1 )
 41         {
 42             a[i-1].Next = NULL;
 43             Num = i-1;
 44             break;
 45         }
 46 
 47 a[i].Addr = a[i-1].NextAddr;
 48         a[i].Data = data[a[i].Addr];
 49         a[i].NextAddr = NextAddress[a[i].Addr];
 50         a[i-1].Next = a+i;
 51 
 52 i++;
 53     }
 54 
 55 iList *p = a; //p指向链表头结点
 56     iList *rp = NULL; //反转链表函数的返回值
 57 
 58 if(K <= Num);
 59     {
 60         for(i = 0;i < (Num/K) ; i++)
 61         {
 62             rp = ListReversing(p,K);
 63             p -> Next = rp;
 64             p -> NextAddr = rp -> Addr;
 65 
 66 int j = 0;
 67             while(j < K)
 68             {
 69                 p = p-> Next;
 70                 j++;
 71             }
 72         }
 73     }
 74 
 75 PrintList(a);
 76     return 0;
 77 }
 78 
 79 iList* ListReversing (iList *p,int K)
 80 {
 81     int count = 1;
 82     iList *new = p ->Next;
 83     iList *old = new ->Next;
 84     iList *temp = NULL ;
 85     while (count < K)
 86     {
 87         temp = old ->Next;
 88         old ->Next = new;
 89         old ->NextAddr = new ->Addr;
 90         new = old;
 91         old = temp;
 92         count++;
 93     }
 94     p ->Next ->Next = old;
 95 
 96 if(old != NULL)
 97     {
 98         p ->Next ->NextAddr = old ->Addr;
 99 
100 }
101     else
102     {
103         p ->Next ->NextAddr = -1;
104     }
105     return new;
106 }
107 void PrintList (iList *a)
108 {
109     iList *p = a;
110     while(p -> Next != NULL){
111         p = p ->Next;
112         if (p->NextAddr != -1 ){
113             //格式输出，%.5意味着如果一个整数不足5位，输出时前面补0 如：22，输出：00022
114             printf("%.5d %d %.5d\n", p->Addr, p->Data, p->NextAddr);
115         }else{
116             //-1不需要以%.5格式输出
117             printf("%.5d %d %d\n", p->Addr, p->Data, p->NextAddr);
118         }
119     }
120 }