## 题目描述

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.

## 解法一

### Python

``````class Solution:
"""
:type S: str
:type K: int
:rtype: str
"""
result = ''
count_k = 0
for letter in reversed(S):
if letter != '-':
result += letter.upper()
count_k += 1
if count_k == K:
result += '-'
count_k = 0
if len(result) != 0 and result[-1] == '-':
result = result[:-1]
return result[::-1]``````

### Java

``````class Solution {
public:
string licenseKeyFormatting(string S, int K) {
string result = "";
int countK = 0;
for (int i = S.length() - 1; i >= 0; i--) {
char letter = S[i];
if (letter != '-') {
result += toupper(letter);
countK++;
if (countK == K) {
result += '-';
countK = 0;
}
}
}
if (result.length() != 0 && result[result.length() - 1] == '-') {
result.pop_back();
}
reverse(result.begin(), result.end());
return result;
}
};``````

### C++

``````public class Solution {
public String licenseKeyFormatting(String S, int K) {
StringBuilder result = new StringBuilder();
int countK = 0;
for (int i = S.length() - 1; i >= 0; i--) {
char letter = S.charAt(i);
if (letter != '-') {
result.append(letter);
countK++;
if (countK == K) {
result.append('-');
countK = 0;
}
}
}
if (result.length() != 0 && result.charAt(result.length() - 1) == '-') {
result.deleteCharAt(result.length() - 1);
}
return result.reverse().toString().toUpperCase();
}
}``````