LeetCode-License Key Formatting

Description：

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:

• The length of string S will not exceed 12,000, and K is a positive integer.
• String S consists only of alphanumerical characters (a-z and/or A-Z and/or 0-9) and dashes(-).
• String S is non-empty.

Java

class Solution {
    public String licenseKeyFormatting(String S, int K) {
        StringBuilder result = new StringBuilder();
        for (int i = S.length() - 1; i >=0;) {
            if (S.charAt(i) == '-') {
                i--;
                continue;
            }
            int k = K;
            while (k > 0) {
                if (i >=0 && S.charAt(i) == '-') {
                    i--;
                    continue;
                }
                if (i < 0) break;
                char ch = Character.isLetter(S.charAt(i)) ? 
                    Character.toUpperCase(S.charAt(i)) : S.charAt(i);
                result.append(ch);
                k--;
                i--;
            }
            result.append('-');
        }
        return result.length() == 0 ? "" : result.reverse().toString().substring(1, result.length());
    }
}