杭电ACM OJ 1020 Encoding 输入一串字符 判断每种字符个数 我犯了个错误定义数组长度的错误

Encoding

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 49076    Accepted Submission(s): 21875


Problem Description

Given a string containing only 'A' - 'Z', we could encode it using the following method: 

1. Each sub-string containing k same characters should be encoded to "kX" where "X" is the only character in this sub-string.

2. If the length of the sub-string is 1, '1' should be ignored.

 

Input

The first line contains an integer N (1 <= N <= 100) which indicates the number of test cases. The next N lines contain N strings. Each string consists of only 'A' - 'Z' and the length is less than 10000.

 

Output

For each test case, output the encoded string in a line.

 

Sample Input

2
ABC
ABBCCC

 

Sample Output

ABC
A2B3C

翻译：不用翻译了
核心：用java的map做，真是简单，当然c，c++做也是一点都不难
错误：我犯了个错误：错误定义了数组长度，想当然地定义了100，然后添加了94个空的‘’到map里去，然后输出的时候就莫名其妙了。

代码:public class Encoding1020 {
char[] mChars;

Encoding1020(char...chars) {
mChars = new char[chars.length];
System.arraycopy(chars, 0, this.mChars, 0, chars.length);
}

private Map<Character, Integer> sort(char[] chars) {
Map<Character, Integer> map = new HashMap<>();

for (int i = 0; i < chars.length; i ++) {
char c = chars[i];
if (!map.containsKey(c)) {
map.put(c, 1);
} else {
int num = map.get(c);
map.put(c, ++num);
}
}

return map;
}

public static void main(String[] args) throws Exception {

Encoding1020 encoding1020 = new Encoding1020('A','B','B','C','C','C');
Map<Character, Integer> map = encoding1020.sort(encoding1020.mChars);
for (Map.Entry<Character, Integer> entry : map.entrySet()) {
char key = entry.getKey();
int value = entry.getValue();
if (value != 1) {
System.out.print(value + "");
}
System.out.print(key + "");
}

}
}