POJ 1002 487-3279

题目链接：http://poj.org/problem?id=1002

题目大意：给一系列的含数字和字母的字符串，其中一些字母对应了一个数字，最终可以将原来的字符串转换成一串数字（电话号码），求出每组号码出现的次数

开始时，做得比较复杂，用了两个 ArrayList,花了很多时间！TLE！后来，使用了 HashMap 得到了优化，终于AC!

 

/**
* @Author：胡家威  
* @CreateTime：2011-4-21 下午08:52:57
* @Description：AC
*/

package ACM.POJ;

import java.util.Arrays;
import java.util.HashMap;
import java.util.Scanner;
import java.util.Set;

public class POJ1002 {
    
     public static void main (String[] args) {
            Scanner scanner = new Scanner(System.in);
            HashMap hs = new HashMap();
            int line = scanner.nextInt();
            boolean nodup = true;
            for (int i=0; i<line; i++){
                char[] chars = new char[8];
                chars[3] = '-';
                int k = 0; 
                String src = scanner.next();
                for (int j = 0; j < src.length(); j++){
                    char c = src.charAt(j);
                    if(c == '-'){
                        continue;
                    }else if(Character.isLetter(c)){
                        if(c=='A' || c=='B' || c=='C'){
                            c = '2';
                        }else if(c=='D' || c=='E' || c=='F'){
                            c = '3';
                        }else if(c=='G' || c=='H' || c=='I'){
                            c = '4';
                        }else if(c=='J' || c=='K' || c=='L'){
                            c = '5';
                        }else if(c=='M' || c=='N' || c=='O'){
                            c = '6';
                        }else if(c=='P' || c=='R' || c=='S'){
                            c = '7';
                        }else if(c=='T' || c=='U' || c=='V'){
                            c = '8';
                        }else if(c=='W' || c=='X' || c=='Y'){
                            c = '9';
                        }
                        if(k == 3){
                            chars[++k] = c;
                            k++;
                        }else{
                            chars[k] = c;
                            k++;
                        }
                    }else if(Character.isDigit(c)){
                        if(k == 3){
                            chars[++k] = c;
                            k++;
                        }else{
                            chars[k] = c;
                            k++;
                        }    
                    }
                }
                String phone = String.valueOf(chars);        
                if(hs.containsKey(phone)){
                    nodup = false;
                    hs.put(phone,(Integer)hs.get(phone)+1);
                }else{
                    hs.put(phone,1);
                }
            }

            if(nodup){
                System.out.println ("No duplicates.");
            }else{
                Set<String> phones = hs.keySet();  //这里 String 很重要！
                String[] str = phones.toArray(new String[phones.size()]);
                Arrays.sort(str);
                for(String s : str){
                    int value = (Integer)hs.get(s);
                    if(value > 1){
                        System.out.println (s+" "+value);
                    }
                }
            }
        }
}