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:
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.

一刷
题解：
StringBuilder insert

class Solution {
    public String licenseKeyFormatting(String S, int K) {
        StringBuilder sb = new StringBuilder();
        for(char c : S.toCharArray()){
            if(c=='-')  continue;
            if(c>='a' && c<='z')    c = Character.toUpperCase(c);
            sb.append(c);
        }
        int insertIndex = sb.length() - K;
        while(insertIndex>0){
            sb.insert(insertIndex,'-');
            insertIndex = insertIndex - K;
        }
        return sb.toString();
    }
}

二刷
由于StringBuilder的insert操作时间复杂度为O(n), 所以采用从头到尾append的方式

class Solution {
    public String licenseKeyFormatting(String s, int k) {
        StringBuilder sb = new StringBuilder();
        char[] cs = s.toCharArray();
        for(char c :cs){
            if(c == '-') continue;
            if(c>='a' && c<='z') c = Character.toUpperCase(c);
            sb.append(c);
        }
        if(sb.length()==0) return "";
        int numDash = sb.length()/k;
        int remain = sb.length() - k*numDash;
        int newLen = sb.length() + numDash;
        StringBuilder res = new StringBuilder();
        int i=0;
        if(remain == 0){
            i = k;
            res.append(sb.substring(0,k));
        }else{
            i = remain;
            res.append(sb.substring(0,remain));
        }
        for(;i