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：

#include<stdio.h>
#define Max 100000
typedef struct Node {
    int Data;
    int Next;
}node;
/* 统计节点的个数 */
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
{
    int cnt=1;
    while((Plist=list[Plist].Next)!=-1)cnt++;
    return cnt;
}
/*输入函数 */
void Input(node *list,int n,int Plist)
{
    int i,addr,Data,Next;
    for (i=1;i<=n;i++)              //利用数组模拟节点！
    {
        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
        list[addr].Data=Data;
        list[addr].Next=Next;  //输入完成后，即形成了一个单向链表
    }
}
/* 逆序函数 */
int Reverse(node *list,int Plist,int cnt,int k)
{
    int prevNode,currNode,nextNode;  //三个基本要素
    prevNode=-1;
    currNode=Plist;
    nextNode=list[currNode].Next;
    int i,j;
    int lasthead,head=-1;
    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
    {
        lasthead=head;   //前一段逆序好的末尾节点
        head=currNode;   //逆序好的该段的末尾节点
    for (i=0;i<k;i++)
    {                                 // 单链表逆序的基本模板！
        list[currNode].Next=prevNode;
        prevNode=currNode;
        currNode=nextNode;
        nextNode=list[nextNode].Next;
    }
    if (j==0)Plist=prevNode;  //整个逆序好的链表的第一个节点
    else list[lasthead].Next=prevNode;  //前一段的尾节点和其下一段的头节点 对上！
    }
    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！
    return Plist;
}
void Printlist(node *list,int Plist)
{
    while((list[Plist].Next)!=-1){
    printf("%05d %d %05d
",Plist,list[Plist].Data,list[Plist].Next);
    Plist=list[Plist].Next;}
    printf("%05d %d %d
",Plist,list[Plist].Data,list[Plist].Next);
}
int main()
{
    int Plist,n,k;
    node list[Max];
    scanf("%d%d%d",&Plist,&n,&k);
    Input(list,n,Plist);
    int cnt;
    cnt=CountNodes(list,Plist);
    Plist=Reverse(list,Plist,cnt,k);
    Printlist(list,Plist);
    return 0;
}