福尔摩斯的约会 (20)

【思路】星期：前面两字符串中第1对相同的大写英文字母

小时：前面两字符串中第2对相同的英文字母，一天的0点到23点由数字0到9、以及大写字母A到N表示

分钟：前面两字符串中第1对相同的字符

import java.io.*;
public class Main{
    public static void main(String[] args)throws IOException{
        BufferedReader bf  = new BufferedReader(new InputStreamReader(System.in));
        String[] day = {"MON", "THE", "WED", "THU","FRI","SAT","SUN"};
         
        char[] ch1 = bf.readLine().toCharArray();
        char[] ch2 = bf.readLine().toCharArray();
        char[] ch3 = bf.readLine().toCharArray();
        char[] ch4 = bf.readLine().toCharArray();
         
        int len1 = ch1.length<ch2.length ? ch1.length: ch2.length;//选取一个长度较小的为单位，方便比较
        boolean flag = true;//没找到
        for(int i=0; i<len1; i++){
            if(ch1[i] == ch2[i]){
                if(flag){
                    //星期
                    if(ch1[i] >= 'A' && ch1[i]<= 'G'){//判断大写字母
                        System.out.print(day[ch1[i] - 'A'] + " ");//day数组的下标就对应ABCDE
                        flag = false;//找到了
                    }
                }else{
                    //小时
                    if(ch1[i] >= '0' && ch1[i]<= '9'){
                        System.out.print("0" +ch1[i] + ":");
                        break;
                    }
                    if(ch1[i] >= 'A' && ch1[i] <= 'N'){
                        System.out.print(10 + ch1[i] -'A' + ":");
                        break;
                    }
                }
            }
        }
        //分钟
        int len2 = ch3.length < ch4.length ?ch3.length : ch4.length;//选取一个长度较小的为单位，方便
        for(int i=0; i<len2 ; i++){
            if(ch3[i] == ch4[i] && ch3[i] >= 'A'){
                if(i<9){
                    System.out.println("0" + i);
                    break;
                }else{
                    System.out.println(i);
                    break;
                }
            }
        }
           
        bf.close();
    }
}