- 例子输入
-
aabbccdd 007799aabbccddeeff113355zz 1234.89898 abcdefabcdefabcdefaaaaaaaaaaaaaabbbbbbbddddddee
- 例子输出
-
abcdabcd 013579abcdefz013579abcdefz <invalid input string> abcdefabcdefabcdefabdeabdeabdabdabdabdabaaaaaaa
Description
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).
Input
Input consists of multiple cases, one case per line. Each case is one string consisting of ASCII characters.
Output
For each case, print exactly one line with the reordered string based on the criteria above.
思路：简单的哈希
import java.text.DecimalFormat;
import java.util.Arrays;
import java.util.Scanner;
import java.lang.String;
import java.lang.Math;
import java.util.HashSet;
/*
class TreeNode
{
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) { val = x; left = null; right = null;}
}
class ListNode
{
int val;
ListNode next;
ListNode(int x){val = x; next = null;}
}
*/
public class Solution {
//static long mode = 1000000007;
/*
public static void permutation(char[] str, HashSet<String> hashset, int start, int end) {
if (start == end) {
hashset.add(new String(str));
//sum++;
}
else {
for (int i = start; i <= end; i++) {
char tmp = str[start];
str[start] = str[i];
str[i] = tmp;
permutation(str, hashset, start+1, end);
tmp = str[start];
str[start] = str[i];
str[i] = tmp;
}
}
}
*/
public static void main(String[] args)
{
//int T ;
Scanner jin = new Scanner(System.in);
//T = jin.nextInt();
while (jin.hasNext()) {
String str = jin.next();
int[] hash_c = new int[75];
for (int i = 0; i < hash_c.length; i++) {
hash_c[i] = 0;
}
int len = str.length();
boolean Isvalid = true;
for (int i = 0; i < len; i++) {
if ((str.charAt(i) >= '0' && str.charAt(i) <= '9') || (str.charAt(i) >= 'a' && str.charAt(i) <= 'z')) {
hash_c[str.charAt(i)-'0']++;
}
else {
Isvalid = false;
break;
}
}
if (!Isvalid) {
System.out.println("<invalid input string>");
continue;
}
StringBuilder strbuild = new StringBuilder();
while (strbuild.length() < len) {
for (int i = 0; i < hash_c.length; i++) {
if (hash_c[i] > 0) {
char ch = (char)(i+'0');
strbuild.append(ch);
hash_c[i]--;
}
}
}
System.out.println(strbuild.toString());
hash_c = null;
}
}
/*
public static double distance(double x, int[] x_array, int[] y_array) {
double dist = 0;
for (int i = 0; i < x_array.length; i++) {
dist += Math.sqrt((x-x_array[i])*(x-x_array[i]) + y_array[i]*y_array[i]);
}
return dist;
}
*/
}