Leetcode 482. License Key Formatting 序列号格式 解题报告

1 解题思想

这题目就是让规整格式的。
输入了一些数字字母还有破折号混合的字符串序列号，和一个K
现在要求将这个序列号，去除原来的破折号格式，整理成每K个字母数字使用一个破折号的格式，如果不能每组都是K个，那么让第一组不是K个，其他组都是K个，最后将小写转变为大写就好

这道题比较简单，直接逆序遍历一下字符串就可以

2 原题

Now you are given a string S, which represents a software license key which we would like to format. The string S is composed of alphanumerical characters and dashes. The dashes split the alphanumerical characters within the string into groups. (i.e. if there are M dashes, the string is split into M+1 groups). The dashes in the given string are possibly misplaced.

We want each group of characters to be of length K (except for possibly the first group, which could be shorter, but still must contain at least one character). To satisfy this requirement, we will reinsert dashes. Additionally, all the lower case letters in the string must be converted to upper case.

So, you are given a non-empty string S, representing a license key to format, and an integer K. And you need to return the license key formatted according to the description above.

Example 1:
Input: S = "2-4A0r7-4k", K = 4

Output: "24A0-R74K"

Explanation: The string S has been split into two parts, each part has 4 characters.
Example 2:
Input: S = "2-4A0r7-4k", K = 3

Output: "24-A0R-74K"

Explanation: The string S has been split into three parts, each part has 3 characters except the first part as it could be shorter as said above.
Note:
The length of string S will not exceed 12,000, and K is 

3 AC解

public class Solution {
    public String licenseKeyFormatting(String S, int K) {
        StringBuilder stringBuilder = new StringBuilder();
        for (int i = S.length() - 1; i >= 0; i--)
            if (S.charAt(i) != '-'){
                if( stringBuilder.length() % (K + 1) == K )
                    stringBuilder.append('-');
                stringBuilder.append(S.charAt(i));
            }
        return stringBuilder.reverse().toString().toUpperCase();
    }
}