Problem

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.

Solution

class Solution {
    public String licenseKeyFormatting(String S, int K) {
        String[] strs = S.toUpperCase().split("-");
        S = "";
        for (String str: strs) {
            S += str;
        }
        int len = S.length();
        StringBuilder sb = new StringBuilder();
        if (len%K != 0) {
            int first = len%K;
            sb.append(S.substring(0, first)+"-");
            S = S.substring(first);
        }
        for (int i = 0; i < len/K; i++) {
            sb.append(S.substring(0, K)+"-");
            S = S.substring(K);
        }
        String res = sb.toString();
        return res.length() > 0 ? res.substring(0, res.length()-1) : "";
    }
}

Solution from @yuxiangmusic

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

linspiration
161 声望53 粉丝