License Key Formatting

Easy

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.

Solution

Kind of verbose (using additional character list, and finding the length of first group before '-') - (13ms

class Solution {
    public String licenseKeyFormatting(String S, int K) {
        int count = 0;
        List<Character> chs = new ArrayList<>();
        for (int i = 0; i < S.length(); i++) {
            if (S.charAt(i) != '-') {
                count++;
                chs.add(S.charAt(i));
            }
        }
        int groupOneLength = count % K;

        StringBuilder sb = new StringBuilder();
        for (int j = 0; j < count; j++) {
            if (j == groupOneLength && groupOneLength > 0) {
                sb.append('-');
            } else if (j > groupOneLength && (j - groupOneLength) % K == 0) {
                sb.append('-');
            }
            sb.append(upperCase(chs.get(j)));
        }
        return sb.toString();
    }

    private char upperCase(char ch) {
        if (ch >= 'a' && ch <= 'z') {
            ch = (char) ('A' + ch - 'a');
        }

        return ch;
    }
}

Preferred (12ms, 92.13%)

用两个index来操作，一个主循环i遍历输入字符串S，一个辅助计数cnt，记录生成的String的段落长度

class Solution {
    public String licenseKeyFormatting(String S, int K) {
        StringBuilder sb = new StringBuilder();
        int cnt = 0;
        for(int i = S.length()-1; i>= 0; i--){
            char c = S.charAt(i);
            if(c == '-') continue;
            if(cnt == K){
                sb.append('-');
                cnt = 0;
            }
            sb.append(Character.toUpperCase(c));
            cnt += 1;
        }
        return sb.reverse().toString();
    }
}

A variation (Using cnt % (K + 1) == K as the timing for adding '-')

class Solution {
    public String licenseKeyFormatting(String S, int K) {
        StringBuilder sb = new StringBuilder();
        int cnt = 0;
        for(int i = S.length()-1; i>= 0; i--){
            char c = S.charAt(i);
            if(c == '-') continue;
            if(cnt % (K + 1) == K){
                sb.append('-');
                cnt += 1;
            }
            sb.append(Character.toUpperCase(c));
            cnt += 1;
        }
        return sb.reverse().toString();
    }
}