[LeetCode][easy]Reformat The String

如果字母的數量和數字的數量之差爲 0 或 1，則可以 reformat。注意如果之差爲 1，那麼多的那種字符應該先寫。

/**
Runtime: 2 ms, faster than 98.95% of Java online submissions for Reformat The String.

Memory Usage: 39.5 MB, less than 100.00% of Java online submissions for Reformat The String.
8、*/
class Solution {
    public String reformat(String s) {
        if (s == null || s.length() == 0) {
            return "";
        }
        int countOfDigit = 0;
        int countOfCharacter = 0;
        char[] digits = new char[s.length()];
        char[] characters = new char[s.length()];
        char[] chars = s.toCharArray();
        for (int i=0; i<chars.length; i++) {
            char c = chars[i];
            if (c >= 'a' && c<= 'z') {
                characters[countOfCharacter] = c;
                countOfCharacter++;
                continue;
            }
            digits[countOfDigit] = c;
            countOfDigit++;
        }
        if (Math.abs(countOfCharacter - countOfDigit) > 1) {
            return "";
        }
            StringBuilder result = new StringBuilder();
            
        if (countOfDigit > countOfCharacter) {
            for (int i=0; i<Math.min(countOfDigit, countOfCharacter); i++) {
                result.append(digits[i]);
                result.append(characters[i]);
            }
            result.append(digits[countOfDigit-1]);
        } else if (countOfDigit < countOfCharacter) {
            for (int i=0; i<Math.min(countOfDigit, countOfCharacter); i++) {
                result.append(characters[i]);
                result.append(digits[i]);
            }
            result.append(characters[countOfCharacter-1]);
        } else {
            for (int i=0; i<Math.min(countOfDigit, countOfCharacter); i++) {
                result.append(characters[i]);
                result.append(digits[i]);
            }
        }
        return result.toString();
        
    }
}