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:
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.
思路：将获得一个表示为字符串S的许可证密钥，该字符串仅包含字母数字字符和短划线。该字符串被N个破折号分成N + 1个组。给定数字K，我们希望重新格式化字符串，使得每个组包含正好的K个字符，但第一个组可能比K短，但仍然必须包含至少一个字符。此外，必须在两个组之间插入短划线，并且所有小写字母都应转换为大写。看完题目第一时间想到的就是使用split分割后使用StringBuilder拼接。
代码如下：
public class LicenseKeyFormatting {
public static String licenseKeyFormatting(String S, int K) {
String[] str = S.split("-");
StringBuilder stringBuilder = new StringBuilder();
for (int i = 0; i < str.length; i++) {
stringBuilder.append(str[i].toUpperCase());
}
if (stringBuilder.length() <= 0) return "";
int oth = stringBuilder.length() % K;
int count = stringBuilder.length() / K;
StringBuilder res = new StringBuilder();
if (oth == 0) {
for (int i = 0; i < count; i++) {
String temp = stringBuilder.substring(i * K, i * K + K);
res.append(temp + "-");
}
} else {
String head = stringBuilder.substring(0, oth);
res.append(head + "-");
StringBuilder stringBuilder1 = stringBuilder.delete(0, oth);
for (int i = 0; i < count; i++) {
String temp = stringBuilder1.substring(i * K, i * K + K);
res.append(temp + "-");
}
}
return res.substring(0, res.length() - 1);
}
}
更简便：
public String licenseKeyFormatting(String S, int K) {
S = S.replaceAll("-", "").toUpperCase();
StringBuilder sb = new StringBuilder(S);
// Starting from the end of sb, and going backwards.
int i = sb.length() - K;
while(i > 0) {
sb.insert(i, '-');
i = i - K;
}
return sb.toString();
}