Description
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.
Solution
Iteration, O(n), S(n)
class Solution {
public String licenseKeyFormatting(String S, int K) {
// store chars in S except dash
StringBuilder sb = new StringBuilder();
for (int i = 0; i < S.length(); ++i) {
if (S.charAt(i) != '-') {
sb.append(S.charAt(i));
}
}
// i should start from sb.length() - K, not S.length() - K!
for (int i = sb.length() - K; i > 0; i -= K) {
sb.insert(i, "-");
}
return sb.toString().toUpperCase();
}
}
网友评论