482.License-Key-Formatting

482. License Key Formatting

题目地址

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.
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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.
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Given a non-empty string S and a number K, format the string according to the rules described above.
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Example 1:
Input: S = "5F3Z-2e-9-w", K = 4
Output: "5F3Z-2E9W"
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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.
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Example 2:
Input: S = "2-5g-3-J", K = 2
Output: "2-5G-3J"
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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.

代码

Approach #1

Time: O(N) && Space: O(1)
class Solution {
public String licenseKeyFormatting(String S, int K) {
StringBuilder sb = new StringBuilder();
int count = 0;
for (int i = S.length() - 1; i >= 0; i--) {
char c = S.charAt(i);
if (c == '-') continue;
if (count % (K + 1) == K) {
sb.append('-');
count++;
}
sb.append(Character.toUpperCase(c));
count++;
}
return sb.reverse().toString();
}
}