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.
//Time: O(n), Space: O(1)    
public String licenseKeyFormatting(String S, int K) {
        if (S == null || S.length() == 0) {
            return S;
        }
        
        S = S.replace("-", "").toUpperCase();
        StringBuilder sb = new StringBuilder();
        int count = 1;//开始从1开始
        
        for (int i = S.length() - 1; i >= 0; i--) {
            sb.insert(0, S.charAt(i));
            
            if (i != 0 && count % K == 0) {//注意:第一位有时虽然count % K == 0但不要加-
                sb.insert(0, '-');
            }
            count++;
        }
        
        return sb.toString();
    }
原文地址:https://www.cnblogs.com/jessie2009/p/9806329.html