482 License Key Formatting

482. License Key Formatting

1. Question

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 containsexactlyK 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.

2. Implementation

class Solution {
    public String licenseKeyFormatting(String S, int K) {
        StringBuilder res = new StringBuilder();

        int n = S.length();

        for (int i = n - 1; i >= 0; i--) {
            if (S.charAt(i) == '-') continue;

            //这里不能 (res.length() % (K + 1)) == 0, 否则刚开始res.length() == 0时，res就会加上'-' 
            if ((res.length() % (K + 1)) == K) {
                res.append('-');
            }
            res.append(S.charAt(i));
        }
        return res.reverse().toString().toUpperCase();
    }
}

3. Time & Space Complexity

时间复杂度O(n), 空间复杂度O(n)