2014微软在线测试-String reorder-WA

题意：

For this question, your program is required to process an input string containing only ASCII characters between ‘0’ and ‘9’, or between ‘a’ and ‘z’ (including ‘0’, ‘9’, ‘a’, ‘z’).

Your program should reorder and split all input string characters into multiple segments, and output all segments as one concatenated string. The following requirements should also be met,
1. Characters in each segment should be in strictly increasing order. For ordering, ‘9’ is larger than ‘0’, ‘a’ is larger than ‘9’, and ‘z’ is larger than ‘a’ (basically following ASCII character order).
2. Characters in the second segment must be the same as or a subset of the first segment; and every following segment must be the same as or a subset of its previous segment.

Your program should output string “<invalid input string>” when the input contains any invalid characters (i.e., outside the '0'-'9' and 'a'-'z' range).

样例输入

aabbccdd007799aabbccddeeff113355zz1234.89898abcdefabcdefabcdefaaaaaaaaaaaaaabbbbbbbddddddee

样例输出

abcdabcd013579abcdefz013579abcdefz<invalid input string>abcdefabcdefabcdefabdeabdeabdabdabdabdabaaaaaaa

代码WA,得了90，不知道哪里错误。

思路，是先计算出每个字符出现的字数，然后根据次数，重组新的字符串输出。

import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner in = new Scanner(System.in);String s = in.next();int length = s.length();if(length<1){    System.out.println("");return;}StringBuilder ret = new StringBuilder();int[] a = new int[256];for(int i=0;i<length;i++){char c = s.charAt(i);if((c>='0' && c<='9') || (c>='a' && c<='z')){a[c]++;}else{System.out.println("<invalid input string>");return;}}boolean b = true;while(b){b = false;for(int i=48;i<123;i++){if(a[i]>0){char c = (char)i;ret.append(c);a[i]--;if(a[i]>0){b=true;}}}}System.out.println(ret.toString());}}