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LicenseKeyFormatting482.java
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/**
* 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.
*/
public class LicenseKeyFormatting482 {
public String licenseKeyFormatting(String S, int K) {
char[] chars = S.toCharArray();
int i = chars.length - 1;
int loop = -1;
StringBuilder sb = new StringBuilder();
while (i >= 0) {
if (chars[i] == '-') {
i--;
continue;
}
sb.append(Character.toUpperCase(chars[i]));
i--;
loop = (loop + 1) % K;
if (loop == K-1) {
sb.append("-");
}
}
sb = sb.reverse();
if (sb.length() > 0 && sb.charAt(0) == '-') sb.deleteCharAt(0);
return sb.toString();
}
}