昨天的JAVA面试题，感觉挺难大家帮忙看看！

//2.2 求解：有一个文本文件，记录了某个学校所有人的姓名、出生日期（假设没有人重名，该校大约有2万人）。

import java.io.BufferedReader;
import java.io.FileInputStream;
import java.io.InputStreamReader;

public class Test1 {
 public static void main(String[] args) {
  //
  int[][] date = new int[12][31];
  int max = 0;
  int month = 0;
  int day = 0;
  int length = 0;
  
  // read file
  try {
   String file = new String("d:\\test\\test.txt");
   BufferedReader bufferedReader =
    new BufferedReader(new InputStreamReader(new FileInputStream(file)));
   String line = bufferedReader.readLine();
   while (line != null) {
    length = line.length();
    month = Integer.parseInt(line.substring(length - 4, length - 2));
    day = Integer.parseInt(line.substring(length - 2, length - 0));
    date[month - 1][day - 1]++;
    line = bufferedReader.readLine();
   }
  } catch (Exception e) {
  }

  // display
  for (int i = 0; i < date.length; i++) {
   for (int j = 0; j < date[i].length; j++) {
    if (date[i][j] > max) {
     max = date[i][j];
     month = i + 1;
     day = j + 1;
    }
   }
  }
  System.out.println("最多人的日期是：" + month + "月" + day + "日");
 }
}

对于第二题这种文本处理工作我觉得用shell更容易，一行完事，不用写这么多代码，其实很多语言工具都可以作同样的事，关键是看实际应用中哪个最合适：
cat data.txt | awk '{print $2}' | sed 's/^.\{4\}//g' | sort | uniq -c | sort -r

测试数据：
a    19800120
b    19920106
c    19970120
d    19150106
e    19380120
f    19990101

结果：人数 生日
      3 0120
      2 0106
      1 0101

万一出现个超长单词：1913个字母，“色氨酸合成酶A蛋白质”（一种含有267种氨基酸酶）的化学名：

MethionylglutaminylarginyltyrosylglutamylserylleucylphenylalanylalanylglutaminylleucyllysylglutamylarginyllysylglutamylglycylalanylphenylalanylvalylprolyphenylalanYlvalythreonylleucylglycylaspartylprolylglycylisoleucylglutamylglutaminylsErylleucyllysylisoleucylaspartylthreonylleucylIsoleucylglutamylalanylglycylalanylasparthlalanylleucylglutamylleucylglycylisoleucylprolylphenylalanylseRylaspartylprolylleucylalanylaspartylglycylpRolylthreOnylisoleucylglutaminylasPfraginylalanylthreonylleucylarfinylalanylphenylalanylalanylalanylglycylvalythreonylprolylalanylglutaminylcysteinylphenylalanylglutamylmethionylleucylalanylleuOylisoleucylarginylglutaminyllysyhistidylprolylthreonylisoleucylprolylisoleucylglycylleucylmethionyltyrosylalanylasparaginylleucylvalylphenylalanylasparaginyllysyglycylisoleucylaspartylglutamylphenylalanylthrosylalanylglutaminylcysteinylglutamyllysylvalylglycylvalylaspartylserylvalylleucylvalylalnylaspartylvalylprolylvalylglUtaminylglutamylserylalanylprolylphenylalanylarginylglutaminylalanylalanylleucylarginylhistidylasparaginyvalylalanylprolylisoleucylprolylisoleucylphenylalanylisoleucylphenylalanylisoleucylcysteinylprolylprolylaspartylalanylaspartylaspartylaspartylleucylleucylarginylglutaminylisoleucylalanylseryltyrosylglycylarginylglycyltyrosylthreonyltyrOsylleucylleucylserylarginylalanylglycylvalylthreonylglycylalanylglutamYlasparainylarginylalanylalanylleucylprolylleucylasparaginylhistidylleucylValylalanyllysylleucyllysylglutamyltyrosylasparaginylalanylalanylprolylprolylleucylglutaminylglgycylphenylalanylglycylisoleucylserylalanylprolylaspartylglutaminylvalyllysylalanylalanylisoleucylaspartylalanylglycylalanylalanylglycylalanylisoleucylserylglycylserylalanylisoleucylvalyllysylisoIeucylisoleucylglutamylglutaminylHistidylasparaginyliSoleucylglutamylprolylglutamyllysylmethionylleucylalanylalanylleucyllysylvalylphenylalanylcalylglutaminylprolylmethionlysylalanylalanylthreonylarginylserine

都转化为质数乘机的话，会是相当相当大的一个数啊！所以在算乘积之前把超长单词去掉是个办法：）

第二道：用map很好解决，key为月日（注意不要年），value为出现次数。变量整个文件，如果map中已经有改日的同学，就加一，如果没有加加入map，value为一。再对map的value排序。
第三道：
    第一步，先对文件转为小写
    第二步，把文件内容转为数组，String s1 = sss.split(" ");
   第三步，对数组中的每个单词排序，排序后的数组s2，如idsa 排序为adis，
   第四步，做了以上准备工作，下面就很好处理了，类似冒泡排序的方法一比较就出来了。
  

第一题
public class Test{
Map<Integer,String[]> map = new HashMap<Integer,String[]>();
public Test(){
String[] s0 = {"0"};
map.put(0, s0);
String[] s1 = {"1"};
map.put(1, s1);
String[] s2 = {"a","b","c"};
map.put(2, s2);
String[] s3 = {"d","e","f"};
map.put(3, s3);
String[] s4 = {"g","h","i"};
map.put(4, s4);
String[] s5 = {"j","k","l"};
map.put(5, s5);
String[] s6 = {"m","n","o"};
map.put(6, s6);
String[] s7 = {"p","q","r","s"};
map.put(7, s7);
String[] s8 = {"t","u","v"};
map.put(8, s8);
String[] s9 = {"w","x","y","z"};
map.put(9, s9);
}
    public static void main(String[] arg){
    new Test().printNum("65967427", "");
    }
   
    public void printNum(String num,String key){
    if(num.length()==0){
    System.out.println(key);
    }else{
    String subnum = num.substring(1, num.length());
    String firstnum = num.substring(0,1);
    for(String s:map.get(Integer.valueOf(firstnum))){
    String newkey = key+s;
    printNum(subnum,newkey);
    }
    }
    }
}

第一题：
import java.util.Scanner;

public class Teltest {
private char[] a0={'0'};
private char[] a1={'1'};
private char[] a2={'a','b','c'};
private char[] a3={'d','e','f'};
private char[] a4={'g','h','i'};
private char[] a5={'j','k','l'};
private char[] a6={'m','n','o'};
private char[] a7={'p','q','r','s'};
private char[] a8={'t','u','v'};
private char[] a9={'w','x','y','z'};
public char[][] m={a0,a1,a2,a3,a4,a5,a6,a7,a8,a9};
public char[][] m1;
public int[] m2;
public char[] m3;
public String istr;
public String pchar;

public void Teltest(){
        Scanner in= new Scanner(System.in);
System.out.print("请输入一串电话号码：");
String istr=in.nextLine();
m1=new char[istr.length()][];
m2=new int[istr.length()];
m3=new char[istr.length()];
pchar="";

for(int i=0;i<istr.length();i++){
m1[i]=m[istr.charAt(i)-48];
pchar=pchar+m1[i][m1[i].length-1];
m2[i]=0;
m3[i]=m1[i][m2[i]];
}
int i=0;
while(!new String(m3).equals(pchar)){
System.out.println(new String(m3));
addchar(i);
}
System.out.println(new String(m3));

}

private void addchar(int i){
if(m2[i]==m1[i].length-1){
m2[i]=0;
i=i+1;
addchar(i);
}else{
m2[i]=m2[i]+1;
}
m3[i]=m1[i][m2[i]];
}


/**
* @param args
* @throws IOException
*/
public static void main(String[] args)  {
// TODO Auto-generated method stub
Teltest t=new Teltest();
t.Teltest();
}

}

用shell是一个不错的思路~哈

这题目要能当场做出来，编程功底的确够厉害的哦！！看样子是专门搞算法的！

第二题：
1、

Map<String,Integer> m=new HashMap<String,Integer> (366);

2、循环读取文本，一行一行的读，截取读出的字符串后四位，记位str
3、

Integer num=m.get(str);
if(num==null){
 m.put(str,1);
}else{
 m.put(str,num+1);
}

4、当文本读取完过后，把m的value中最大数对应key的取出来，key就是举行生日舞会的日期了 ，赶快去参加吧

  和我的想法一样啊