LeetCode #482 - License Key Formatting-优快云博客

本文介绍了如何将给定的包含字母和数字的字符串按照特定规则进行格式化，并且将小写字母转换为大写。

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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.

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.

将字符串按照每k个字符隔开，但是第一部分不一定为k个字符。

class Solution {
public:
    string licenseKeyFormatting(string S, int K) {
        istringstream iss(S);
        string temp;
        string s;
        while(getline(iss,temp,'-')) s+=temp;
        int n=s.size();
        for(int i=0;i<n;i++)
            if(s[i]>='a'&&s[i]<='z') s[i]=s[i]-'a'+'A';
        
        if(n<K) return s;
        string result;
        int i=n;
        while(i>K)
        {
            result="-"+s.substr(i-K,K)+result;
            i-=K;
        }
        result=s.substr(0,i)+result;
        return result;
    }
};