Description

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

給一個String，表示車牌，K表示處理過之後每一段文字的長度，除第一段可以不滿K個，之後的都要K個的形式。

Using a StringBuilder to construct new string. Form the tail of the string, we start iteration. The ‘-’ always shows at k+1 position. So we use the length of sb % (k + 1) to determine whether we should add a ‘-’.

After the iteration, reverse sb and set it to upper case would sovle the problem.

Code

class Solution {
    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();
    }
}

Time Complexity: O(n) Space Complexity: O(n)

Review