PAT 02-线性结构1. Reversing Linked List

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 }

ListReversing函数不仅要交换指针，还要交换addr.

测试点一共有7个： L 代表单链表节点数，因为构成单链表的节点不一定都在输入的N个节点中，即：L<=N;

case 0：L = N 有节点 Address = 99999

case 1: L = mK, L = N, (m = 2, 3, 4,...) 有节点 Address = 99999

case 2: K = N, L = N 有节点 Address = 99999

case 3: K = 1, L = mK 有节点 Address = 99999

case 4: K = 1, L = N = 1 （很简单的一个测试点）

case 5: K != 1, L % K = (K-1) （节点数很多，如果建链表的复杂度是O(n*n), 超时的可能性很大）

case 6: L > N （有多余节点） 有节点Address = 99999

要考虑的细节：K=1不反转，K=L 全反转，L%K == 0, 每段都反转，L%k = (K-1),多余的节点不反转。L < N ，有多余节点的情况。

thanks:

PAT02-1Reversing Linked List (25) 单链表逆序

单链表反转/逆序的两种方法