PAT1074 Reversing Linked List (25)詳細題解

 

02-1. Reversing Linked List (25)

時間限制

400 ms

內存限制

65536 kB

代碼長度限制

8000 B

判題程序

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

 

這題用二維數組高維代表首地址啊address，低維[address][0]儲存data,[address][1]則儲存next，犧牲空間換時間。

再使用維數組list進行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;
    }

先輸入node值。如圖

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

 

 

 

值得說明的是：不需要按照順序輸入，到數組中后自然會對應數組的下標值即address。

因此上表中按照數組順序排序，無初始值的數組下標省略不列。

下面對list賦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]儲存的是st的address，list[1]儲存的是st->next的地址，以此類推。直到遇到address為-1.

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

進行反轉，使用algorithm的reverse函數。

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

最后進行輸出，注意的是以int形式儲存的，如address中的00000實際儲存位置是0，故輸出時注意是要用五位數輸出。

完整AC代碼如下：

#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\n", list[i], node[list[i]][0], list[i+1]);
    }
    printf("%05d %d -1\n", list[i], node[list[i]][0]);
}