# LeetCode: 482. License Key Formatting¶

## 1、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.

## 2、解题思路¶

​ 一开始没有理解题意，实际上，他的意思是，假如使用3个字符一组，那么所有的字符，除了第一个，其他的都是3个字符

char* licenseKeyFormatting(char* S, int K) {

int length = strlen(S);

int length_temp = length;

int result_pos = length + length / K;

char *temp = S + length - 1;

char *result = (char *) calloc(length + length / K, sizeof(char));

int count = K;
while (length_temp-- > 0) {
if (*temp != '-') {
if (count == 0) {
result[result_pos--] = '-';
count = K;
}
if (*temp >= 'a' && *temp <= 'z') {
result[result_pos--] = *temp - 32;
} else {
result[result_pos--] = *temp;
}

count--;
}

temp--;

}

return &result[result_pos+1];

}