1006 Sign In and Sign Out

解法一：字符串处理

简单的字符串处理问题，根据时间格式统一转换成以秒为单位的时间，然后统计最大值和最小值即可。

import java.io.*;
public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        int N = Integer.parseInt(in.readLine());
        String[] input;
        int signInTime = 0x3f3f3f3f, signOutTime = -1;
        String signInId = "", signOutId = "";
        for (int i = 0; i < N; ++i) {
            input = in.readLine().split(" ");
            int signIn = parseTime(input[1]);
            int signOut = parseTime(input[2]);
            if (signIn < signInTime) {
                signInTime = signIn;
                signInId = input[0];
            }
            if (signOut > signOutTime) {
                signOutTime = signOut;
                signOutId = input[0];
            }
        }
        out.println(signInId + " " + signOutId);
        out.flush();
    }
    private static int parseTime(String time) {
        String[] elems = time.split(":");
        return Integer.parseInt(elems[0]) * 3600 +
                Integer.parseInt(elems[1]) * 60 +
                Integer.parseInt(elems[2]);
    }
}