PAT1074 Reversing Linked List (25)详细题解

02-1. Reversing Linked List (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

8000 B

判题程序

Standard

作者

CHEN, Yue

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

这题用二维数组高维代表首地址啊address，低维[address][0]储存data,[address][1]则储存next，牺牲空间换时间。

再使用维数组list进行reverse.

int list[100010];
int node[100010][2];

    int st,num,r;
    cin>>st>>num>>r;
    int address,data,next,i;
    for(i=0;i<num;i++)
    {
        cin>>address>>data>>next;
        node[address][0]=data;
        node[address][1]=next;
    }

先输入node值。如图

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

值得说明的是：不需要按照顺序输入，到数组中后自然会对应数组的下标值即address。

因此上表中按照数组顺序排序，无初始值的数组下标省略不列。

下面对list赋address.

int m=0,n=st;
while(n!=-1)
{
    list[m++]=n;
    n=node[n][1];
}

list	0	1	2	3	4	5
address	00100	12309	33218	00000	99999	68237

其中list[0]储存的是st的address，list[1]储存的是st->next的地址，以此类推。直到遇到address为-1.

i=0;
while(i+r<=m)
{
    reverse(list+i,list+i+r);
    i=i+r;
}

进行反转，使用algorithm的reverse函数。

for (i = 0; i < m-1; i++)
{
    printf("%05d %d %05d
", list[i], node[list[i]][0], list[i+1]);
}
printf("%05d %d -1
", list[i], node[list[i]][0]);

最后进行输出，注意的是以int形式储存的，如address中的00000实际储存位置是0，故输出时注意是要用五位数输出。

完整AC代码如下：

#include<iostream>
#include<algorithm>
using namespace std;
int list[100010];
int node[100010][2];
int main()
{
    freopen("F:\in1.txt","r",stdin);
    int st,num,r;
    cin>>st>>num>>r;
    int address,data,next,i;
    for(i=0;i<num;i++)
    {
        cin>>address>>data>>next;
        node[address][0]=data;
        node[address][1]=next;
    }
    int m=0,n=st;
    while(n!=-1)
    {
        list[m++]=n;
        n=node[n][1];
    }
    i=0;
    while(i+r<=m)
    {
        reverse(list+i,list+i+r);
        i=i+r;
    }
    for (i = 0; i < m-1; i++)
    {
        printf("%05d %d %05d
", list[i], node[list[i]][0], list[i+1]);
    }
    printf("%05d %d -1
", list[i], node[list[i]][0]);
}