02-线性结构3 Reversing Linked List

02-线性结构3 Reversing Linked List

分数 25
作者 陈越
单位 浙江大学

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
1
2
3
4
5
6
7

Sample Output:

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

代码长度限制
16 KB
时间限制
400 ms
内存限制
64 MB
C++ (g++)

思路：

用Data数组存数值，用Next数组存下一节点的地址，两个都以当前节点的地址为下标，同时记录了值，地址值，下一节点的地址。
用OriAdr数组用来存输入数据的地址值
开一个vector容器用来存反转后节点的地址，然后根据这个地址下标，输出地址、值、下一节点的地址。
Read(int *Data,int * Next, int N) 读入数据
Print_Format (int num) 格式化输出五位数地址

Reverse(int * Data,int * Next,int first,int K)
逆置链表
首先记录节点的个数
然后根据个数确定够不够K，逆不逆置
如果不能逆置的话，直接将剩下的节点地址值存入Ans中
如果能逆置的话，将这一组节点的地址值倒序压入Ans中
如果遍历完整个链表节点的话，退出循环。
输出逆置后的链表
注意结尾不能有多余空格
主函数就很简单。

AC代码：

#include
using namespace std;
#define maxsize 100005
int Data[maxsize];
int Next[maxsize];
int OriAdr[maxsize];//输入顺序的地址
vector<int> Ans;//反转后节点的地址
void Read(int *Data,int *Next, int N)
{
    for(int i=0;i<N;i++)
    {
        int address,data,next;
        cin>>address>>data>>next;
        Data[address]=data;//以地址为下标
        Next[address]=next;
    }
}

void Print_Format(int num)
{
    //格式化输出五位地址数
    if(num==-1) {cout<<-1;return;}
    int d=10000;
    for(int i=0;i<5;i++)
    {
        int temp=num/d;
        while(temp/10!=0) temp=temp%10;
        cout<<temp;
        d/=10;
    }
}
void Reverse(int* Data,int* Next,int first,int K)
{
    int temp=first;
    int cnt=0;
    //计算有效节点，题目所给有些节点是多余的
    while(temp!=-1)
    {
        cnt++;
        OriAdr[cnt]=temp;
        temp =Next[temp];//按照链表地址顺序索引
    }
    for(int i=0;;i+=K)
    {
        //若剩余节点不足k，则将余下节点按照顺序加入Ans中
        if(i+K>cnt)
        {
            i++;//从第i+1个节点开始
            while(i<=cnt)
            {
                Ans.push_back(OriAdr[i]);
                i++;
            }
            break;
        }
        int j=i+K;
        while(j>i)
        {
            Ans.push_back(OriAdr[j]);
            j--;
        }
        if(i+K==cnt) break;
    }
    for(auto it=Ans.begin();it!=Ans.end();it++)
    {
        Print_Format(*it);
        cout<<" "<<Data[*it]<<" ";
        if(it+1==Ans.end()) cout<<-1;
        else
        {
            Print_Format(*(it+1)); cout<<endl;
        }
    }
}
int main()
{
    int first,N,K;//首地址，节点个数，反转个数
    cin>>first>>N>>K;
    Read(Data,Next,N);
    Reverse(Data,Next,first,K);
    return 0;
}
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