对称排序

描述

In your job at Albatross Circus Management (yes, it's run by a bunch of clowns), you have just finished writing a program whose output is a list of names in nondescending order by length (so that each name is at least as long as the one preceding it). However, your boss does not like the way the output looks, and instead wants the output to appear more symmetric, with the shorter strings at the top and bottom and the longer strings in the middle. His rule is that each pair of names belongs on opposite ends of the list, and the first name in the pair is always in the top part of the list. In the first example set below, Bo and Pat are the first pair, Jean and Kevin the second pair, etc.

输入

The input consists of one or more sets of strings, followed by a final line containing only the value 0. Each set starts with a line containing an integer, n, which is the number of strings in the set, followed by n strings, one per line, NOT SORTED. None of the strings contain spaces. There is at least one and no more than 15 strings per set. Each string is at most 25 characters long.

输出

For each input set print "SET n" on a line, where n starts at 1, followed by the output set as shown in the sample output.
If length of two strings is equal,arrange them as the original order.(HINT: StableSort recommanded)

样例输入

7BoPatJeanKevinClaudeWilliamMarybeth6JimBenZoeJoeyFrederickAnnabelle5JohnBillFranStanCece0

样例输出

SET 1BoJeanClaudeMarybethWilliamKevinPatSET 2JimZoeFrederickAnnabelleJoeyBenSET 3JohnFranCeceStan
Bill

import java.util.*;  public class Main {     static Comparator<String> com=new Comparator<String>() {@Overridepublic int compare(String o1, String o2) {return o1.length()-o2.length();}};static int cnt=0;    public static void main(String[] args) {          Scanner in= new Scanner(System.in);           while(in.hasNext()){          int m=in.nextInt();          if(m==0)         break;         String[] s=new String[m];         String[] s1=new String[m];         int pos=0;         for (int i = 0; i <m; i++) s[i]=in.next();         Arrays.sort(s, com);         for (int i = 0; i < s1.length; i++) {     if(i%2==0)                s1[pos++]=s[i];elses1[s1.length-pos]=s[i];}     System.out.println("SET"+" "+(cnt+1));         for (int i = 0; i < s1.length; i++)              System.out.println(s1[i]);         cnt++;        }     }      }

反思：真是绝了，我英语真是差啊，题目意思没有理解透彻，而且题目所给的字符串都是已经排序好了的，所以就是先对字符串按照长度排序再进行对称输出就好了，就很简单了