Reversing Linked List

原题连接：https://www.patest.cn/contests/pat-a-practise/1074

题目：

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

我的代码：

 1 #include<stdio.h>
 2 #define Max 100000
 3 typedef struct Node {
 4     int Data;
 5     int Next;
 6 }node;
 7 /* 统计节点的个数 */
 8 int CountNodes(node *list,int Plist)   //因为题目中有多余的节点，故要统计节点个数
 9 {
10     int cnt=1;
11     while((Plist=list[Plist].Next)!=-1)cnt++;
12     return cnt;
13 }
14 /*输入函数 */
15 void Input(node *list,int n,int Plist)
16 {
17     int i,addr,Data,Next;
18     for (i=1;i<=n;i++)              //利用数组模拟节点！
19     {
20         scanf("%d%d%d",&addr,&Data,&Next);  //不必按照顺序输入，是因为只要有“头节点”，就可以找到其他元素了
21         list[addr].Data=Data;
22         list[addr].Next=Next;  //输入完成后，即形成了一个单向链表
23     }
24 }
25 /* 逆序函数 */
26 int Reverse(node *list,int Plist,int cnt,int k)
27 {
28     int prevNode,currNode,nextNode;  //三个基本要素
29     prevNode=-1;
30     currNode=Plist;
31     nextNode=list[currNode].Next;
32     int i,j;
33     int lasthead,head=-1;
34     for (j=0;j<cnt/k;j++)  //分为 cnt/k 段进行逆序，每段k个节点 。后面的余数不必逆序
35     {
36         lasthead=head;   //前一段逆序好的末尾节点
37         head=currNode;   //逆序好的该段的末尾节点
38     for (i=0;i<k;i++)
39     {                                 // 单链表逆序的基本模板！
40         list[currNode].Next=prevNode;
41         prevNode=currNode;
42         currNode=nextNode;
43         nextNode=list[nextNode].Next;
44     }
45     if (j==0)Plist=prevNode;  //整个逆序好的链表的第一个节点
46     else list[lasthead].Next=prevNode;  //前一段的尾节点和其下一段的头节点 对上！
47     }
48     list[head].Next=currNode;  //进行逆序的最后一段的末尾节点和剩余没有进行逆序的段的头节点 对上！
49     return Plist;
50 }
51 void Printlist(node *list,int Plist)
52 {
53     while((list[Plist].Next)!=-1){
54     printf("%05d %d %05d
",Plist,list[Plist].Data,list[Plist].Next);
55     Plist=list[Plist].Next;}
56     printf("%05d %d %d
",Plist,list[Plist].Data,list[Plist].Next);
57 }
58 int main()
59 {
60     int Plist,n,k;
61     node list[Max];
62     scanf("%d%d%d",&Plist,&n,&k);
63     Input(list,n,Plist);
64     int cnt;
65     cnt=CountNodes(list,Plist);
66     Plist=Reverse(list,Plist,cnt,k);
67     Printlist(list,Plist);
68     return 0;
69 }