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.

程序如下所示：

class Solution {
    public String licenseKeyFormatting(String S, int K) {
        StringBuilder sb = new StringBuilder();
        int len = S.length();
        int size = 0;
        int letters = 0;
        for (int i = 0; i < len; ++ i){
            if (S.charAt(i) != '-'){
                letters ++;
            }
        }
        boolean tag = true;
        for (int i = 0; i < len; ++ i){
            if (S.charAt(i) != '-'){
                size ++;
                sb.append(S.charAt(i));
            }
            if (size >= K){
                sb.append('-');
                size = 0;
            }
            if (tag && size != 0 && size == (letters%K)){
                sb.append('-');
                size = 0;
                tag = false;
            }
        }
        return sb.length() != 0 ? sb.substring(0, sb.length() - 1).toUpperCase():"";
    }
}

本题还可以从后往前处理，这样就可以省去第一个部分的处理。