【PAT】【Advanced Level】1074. Reversing Linked List (25)

1074. Reversing Linked List (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 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 400000 4 9999900100 1 1230968237 6 -133218 3 0000099999 5 6823712309 2 33218

Sample Output:

00000 4 3321833218 3 1230912309 2 0010000100 1 9999999999 5 6823768237 6 -1

原题链接：

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

https://www.nowcoder.com/pat/5/problem/4033

思路：

读入，用map映射，整理出原顺序的列表

然后按块翻转，注意两块之间的处理

CODE：

#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<cstdlib>#include<map>#define N 100010using namespace std;map<int,int> ma;typedef struct S{int addr;int nxt;int val;};S pre[N];S no[N];int main(){char bg[5];int n,m;int head;scanf("%s %d %d",bg,&n,&m);head=atoi(bg);for (int i=0;i<n;i++){char* s1=(char*)malloc(5*sizeof(char));char* s2=(char*)malloc(5*sizeof(char));scanf("%s %d %s",s1,&pre[i].val,s2);pre[i].addr=atoi(s1);pre[i].nxt=atoi(s2);ma[pre[i].addr]=i;}int len=0;int now=head;while (now!=-1){no[len]=pre[ma[now]];now=pre[ma[now]].nxt;len++;}no[len].addr=-1;for (int i=0;i<(len/m)*m;i+=m){for (int j=i+m-1;j>i;j--){no[j].nxt=no[j-1].addr;printf("%05d %d %05d\n",no[j].addr,no[j].val,no[j].nxt);}if (i+2*m>len)no[i].nxt=no[i+m].addr;elseno[i].nxt=no[i+2*m-1].addr;if (no[i].nxt!=-1)printf("%05d %d %05d\n",no[i].addr,no[i].val,no[i].nxt);elseprintf("%05d %d -1",no[i].addr,no[i].val);}for (int i=(len/m)*m;i<len;i++){if (no[i].nxt!=-1)printf("%05d %d %05d\n",no[i].addr,no[i].val,no[i].nxt);elseprintf("%05d %d -1",no[i].addr,no[i].val);}return 0;}