482. License Key Formatting

最新推荐文章于 2024-07-24 17:52:56 发布

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最新推荐文章于 2024-07-24 17:52:56 发布

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分类专栏： leetcode 文章标签： string

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本文介绍了一种算法，用于按照特定格式重新排列软件许可密钥。该算法首先去除原始密钥中的分隔符并统一转换为大写字母，然后按照指定长度K重新插入分隔符。

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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 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.

题目是将一个序列号按照一定格式输出。首先要将原序列号中的分隔符‘-’删除，并把字母都换成大写。然后每隔K个加一个分隔符‘-’，分隔成若干个子序列，注意第一个子序列字符个数可以小于或等于K个。

代码：

class Solution 
{
public:
	string licenseKeyFormatting(string S, int K) 
	{
		string res, tmp;
		if(S.empty()) return res; 
		for(auto &c:S)
		{
			if(c != '-')
			{
				tmp += toupper(c);
			}
		}
		int cnt = tmp.size(), i = cnt % K;
		res = tmp.substr(0, i);
		for(int j = 0; j < cnt / K; ++j)
		{
			if(i != 0) res += '-';
			res += tmp.substr(i, K);
			i += K;
		}
		return res;
	}
};