1006. Sign In and Sign Out (25)-----Java

xiaoxiao2021-02-28  7

题目内容 At the beginning of every day, the first person who signs in the computer room will unlock the door, and the last one who signs out will lock the door. Given the records of signing in’s and out’s, you are supposed to find the ones who have unlocked and locked the door on that day.

Input Specification:

Each input file contains one test case. Each case contains the records for one day. The case starts with a positive integer M, which is the total number of records, followed by M lines, each in the format:

ID_number Sign_in_time Sign_out_time where times are given in the format HH:MM:SS, and ID number is a string with no more than 15 characters.

Output Specification:

For each test case, output in one line the ID numbers of the persons who have unlocked and locked the door on that day. The two ID numbers must be separated by one space.

Note: It is guaranteed that the records are consistent. That is, the sign in time must be earlier than the sign out time for each person, and there are no two persons sign in or out at the same moment.

Sample Input: 3 CS301111 15:30:28 17:00:10 SC3021234 08:00:00 11:25:25 CS301133 21:45:00 21:58:40 Sample Output: SC3021234 CS301133 题目分析 1.找出进入最小的时间所对应的名字和出去最大的时间所对应的名字。 2.不需要保存输入的数据 代码

//不保存 import java.util.Scanner; public class Main2 { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); String name1="",name2=""; int min = 24*3600+60*60+60,max = -1,index1=-1,index2=-1; for (int i = 0; i <n; i++) { String name = sc.next(); String sign_in = sc.next(); String sign_out = sc.next(); String[] in = sign_in.split(":"); String[] out = sign_out.split(":"); int signin = Integer.valueOf(in[0])*3600 + Integer.valueOf(in[1])*60+ Integer.valueOf(in[2]); int signout = Integer.valueOf(out[0])*3600 + Integer.valueOf(out[1])*60+ Integer.valueOf(out[2]); if(min>signin){ min = signin; name1 = name; } if(max<signout){ max = signout; name2 = name; } } System.out.println(name1+" "+name2); } } //保存 import java.util.Scanner; class persons{ public String name; int sign_in; int sign_out; persons(){ } persons(String name,int in,int out){ this.name = name; sign_in = in; sign_out = out; } } public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); persons people[] = new persons[n]; for (int i = 0; i < people.length; i++) { String name = sc.next(); String sign_in = sc.next(); String sign_out = sc.next(); String[] in = sign_in.split(":"); String[] out = sign_out.split(":"); int signin = Integer.valueOf(in[0])*3600 + Integer.valueOf(in[1])*60+ Integer.valueOf(in[2]); int signout = Integer.valueOf(out[0])*3600 + Integer.valueOf(out[1])*60+ Integer.valueOf(out[2]); people[i] = new persons(name, signin, signout); } int min = 24*3600+60*60+60,max = -1,index1=-1,index2=-1; for (int i = 0; i < people.length; i++) { if(min>people[i].sign_in){ min = people[i].sign_in; index1 = i; } if(max<people[i].sign_out){ max = people[i].sign_out; index2 = i; } } System.out.println(people[index1].name+" "+people[index2].name); } }
转载请注明原文地址: https://www.6miu.com/read-200242.html

最新回复(0)