题目链接:https://www.nowcoder.com/practice/0f64518fea254c0187ccf0ea05019672?tpId=40&tqId=21363&tPage=2&rp=2&ru=/ta/kaoyan&qru=/ta/kaoyan/question-ranking
题目描述
有一个网络日志,记录了网络中计算任务的执行情况,每个计算任务对应一条如下形式的日志记录: “hs_10000_p”是计算任务的名称, “2007-01-17 19:22:53,315”是计算任务开始执行的时间“年-月-日 时:分:秒,毫秒”, “253.035(s)”是计算任务消耗的时间(以秒计) hs_10000_p 2007-01-17 19:22:53,315 253.035(s) 请你写一个程序,对日志中记录计算任务进行排序。 时间消耗少的计算任务排在前面,时间消耗多的计算任务排在后面。 如果两个计算任务消耗的时间相同,则将开始执行时间早的计算任务排在前面。
输入描述:
日志中每个记录是一个字符串,每个字符串占一行。最后一行为空行,表示日志结束。日志中最多可能有10000条记录。 计算任务名称的长度不超过10,开始执行时间的格式是YYYY-MM-DD HH:MM:SS,MMM,消耗时间小数点后有三位数字。 计算任务名称与任务开始时间、消耗时间之间以一个或多个空格隔开,行首和行尾可能有多余的空格。
输出描述:
排序好的日志记录。每个记录的字符串各占一行。 输入的格式与输入保持一致,输入包括几个空格,你的输出中也应该包含同样多的空格。
示例1
输入
复制hs_10000_p 2007-01-17 19:22:53,315 253.035(s) hs_10001_p 2007-01-17 19:22:53,315 253.846(s) hs_10002_m 2007-01-17 19:22:53,315 129.574(s) hs_10002_p 2007-01-17 19:22:53,315 262.531(s) hs_10003_m 2007-01-17 19:22:53,318 126.622(s) hs_10003_p 2007-01-17 19:22:53,318 136.962(s) hs_10005_m 2007-01-17 19:22:53,318 130.487(s) hs_10005_p 2007-01-17 19:22:53,318 253.035(s) hs_10006_m 2007-01-17 19:22:53,318 248.548(s) hs_10006_p 2007-01-17 19:25:23,367 3146.827(s)
输出
复制hs_10003_m 2007-01-17 19:22:53,318 126.622(s) hs_10002_m 2007-01-17 19:22:53,315 129.574(s) hs_10005_m 2007-01-17 19:22:53,318 130.487(s) hs_10003_p 2007-01-17 19:22:53,318 136.962(s) hs_10006_m 2007-01-17 19:22:53,318 248.548(s) hs_10000_p 2007-01-17 19:22:53,315 253.035(s) hs_10005_p 2007-01-17 19:22:53,318 253.035(s) hs_10001_p 2007-01-17 19:22:53,315 253.846(s) hs_10002_p 2007-01-17 19:22:53,315 262.531(s) hs_10006_p 2007-01-17 19:25:23,367 3146.827(s)
代码
#include<iostream> #include<string> #include<algorithm> using namespace std; struct Log{ char str[100]; int year; int month; int day; int hour; int minutes; int seconds; int ms; double times; }log[10001]; bool cmp(Log L1,Log L2) { if(L1.times!=L2.times) return L1.times<L2.times; else if(L1.year != L2.year) return L1.year < L2.year; else if(L1.month != L2.month) return L1.month < L2.month; else if(L1.day != L2.day) return L1.day < L2.day; else if(L1.hour != L2.hour) return L1.hour < L2.hour; else if(L1.minutes != L2.minutes) return L1.minutes < L2.minutes; else if(L1.seconds != L2.seconds) return L1.seconds < L2.seconds; else return L1.ms < L2.ms; } int main() { int i = 0; while(gets(log[i].str)) { if(strcmp(log[i].str,"")==0) break; sscanf(log[i].str,"%*s%4d-%2d%*c%2d %2d:%2d%*c%2d%*c%3d %lf%*s",&log[i].year,&log[i].month,&log[i].day,&log[i].hour,&log[i].minutes,&log[i].seconds,&log[i].ms,&log[i].times); i++; } cout<<i<<endl; sort(log,log+i,cmp); for(int j = 0;j<i;j++) { cout<<log[j].str <<endl; } return 0; }
运行时间:13ms
占用内存:1388k
由于要记录前后中间的空格,所以用gets()函数会比较方便,同时利用sscanf 将数据提取;利用sort函数实现排序并重写cmp排序方法
或者使用qsort函数来实现
#include<iostream> #include<string> #include<stdio.h> #include<stdlib.h> #include<string.h> using namespace std; struct Log{ char str[100]; int year; int month; int day; int hour; int minutes; int seconds; int ms; double times; }log[10001]; int cmp(const void *a,const void *b) { struct Log *L1 = (Log *)a; struct Log *L2 = (Log *)b; if(L1->times!=L2->times) return L1->times>L2->times?1:-1; else if(L1->year != L2->year) return L1->year - L2->year; else if(L1->month != L2->month) return L1->month - L2->month; else if(L1->day != L2->day) return L1->day - L2->day; else if(L1->hour != L2->hour) return L1->hour - L2->hour; else if(L1->minutes != L2->minutes) return L1->minutes - L2->minutes; else if(L1->seconds != L2->seconds) return L1->seconds - L2->seconds; else return L1->ms - L2->ms; } int main() { int i = 0; while(gets(log[i].str)) { if(strcmp(log[i].str,"")==0) break; sscanf(log[i].str,"%*s%4d-%2d%*c%2d %2d:%2d%*c%2d%*c%3d %lf%*s",&log[i].year,&log[i].month,&log[i].day,&log[i].hour,&log[i].minutes,&log[i].seconds,&log[i].ms,&log[i].times); i++; } qsort(log,i,sizeof(log[0]),cmp); //sort(log,log+i,cmp); for(int j = 0;j<i;j++) { cout<<log[j].str <<endl; } return 0; }