482. License Key Formatting

You are given a license key represented as a string S which consists only alphanumeric character and dashes. The string is separated into N+1 groups by N dashes.

Given a number K, we would want to reformat the strings such that each group contains exactly K characters, except for the first group which could be shorter than K, but still must contain at least one character. Furthermore, there must be a dash inserted between two groups and all lowercase letters should be converted to uppercase.

Given a non-empty string S and a number K, format the string according to the rules described above.

Example 1:

Input: S = "5F3Z-2e-9-w", K = 4

Output: "5F3Z-2E9W"

Explanation: The string S has been split into two parts, each part has 4 characters.
Note that the two extra dashes are not needed and can be removed.

Example 2:

Input: S = "2-5g-3-J", K = 2

Output: "2-5G-3J"

Explanation: The string S has been split into three parts, each part has 2 characters except the first part as it could be shorter as mentioned above.

Note:

1. The length of string S will not exceed 12,000, and K is a positive integer.

2. String S consists only of alphanumerical characters (a-z and/or A-Z and/or 0-9) and dashes(-).

3. String S is non-empty.

解题要点：

从字符串尾开始处理，只用关心非“-”的字符，当检测到字符长度满足K时，手动加上“-”和当前字符。返回这个字符串的反转串。

class Solution {
    public String licenseKeyFormatting(String S, int K) {
        StringBuilder res = new StringBuilder();
        for(int i = S.length() - 1; i >= 0; i--){
            char s = S.charAt(i);
            if(s != '-'){
                String temp = res.length() % (K + 1) == K ? "-" : "";
                res.append(temp).append(s);                
            }
        }
        return res.reverse().toString().toUpperCase();
    }
}