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:
struct Solution;
impl Solution {
fn license_key_formatting(s: String, k: i32) -> String {
let mut res: Vec<char> = vec![];
let mut i = 0;
for c in s.chars().rev() {
if c != '-' {
res.push(c);
i += 1;
if i == k {
i = 0;
res.push('-');
}
}
}
if let Some(&'-') = res.last() {
res.pop();
}
res.iter().rev().collect::<String>().to_ascii_uppercase()
}
}
#[test]
fn test() {
let s = "5F3Z-2e-9-w".to_string();
let k = 4;
let o = "5F3Z-2E9W".to_string();
assert_eq!(Solution::license_key_formatting(s, k), o);
let s = "2-5g-3-J".to_string();
let k = 2;
let o = "2-5G-3J".to_string();
assert_eq!(Solution::license_key_formatting(s, k), o);
}