POJ 1002 487-3279
2011-07-15 21:12
477 查看
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!
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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); } } } } }
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