leetcode 482. License Key Formatting

我就是我 2022-06-03 01:10 63阅读 0赞

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.

题意很简单，注意所有的字符都要变成大写，然后按照破折直接拼接即可

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <cmath>
    
    using namespace std;
    
    class Solution
    {
    public:
        string licenseKeyFormatting(string S, int K)
        {
            if (S.size()<=0)
                return S;
    
            string res, tmp;
            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;
        }
    };