482. License Key Formatting(easy)

Easy

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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:

1. The length of string S will not exceed 12,000, and K is a positive integer.
2. String S consists only of alphanumerical characters (a-z and/or A-Z and/or 0-9) and dashes(-).
3. String S is non-empty.

 

C++:

/*
 * @Autor: SourDumplings
 * @Date: 2019-09-27 13:48:55
 * @Link: https://github.com/SourDumplings/
 * @Email: [email protected]
 * @Description: https://leetcode.com/problems/license-key-formatting/
 */

class Solution
{
public:
    string licenseKeyFormatting(string S, int K)
    {
        string s;
        int L = S.length();
        for (int i = 0; i < L; ++i)
        {
            char c = S[i];
            if (isalnum(c))
            {
                s += toupper(c);
            }
        }
        int l = s.length();
        int first = l % K;
        bool isFirst = (first != 0);
        string res;
        for (int i = 0; i < l; i++)
        {
            res += s[i];
            if (isFirst)
            {
                if (i == first - 1 && K < l)
                {
                    res += '-';
                    isFirst = false;
                }
            }
            else if ((i - first + 1) % K == 0 && i != l - 1)
            {
                res += '-';
            }
        }

        return res;
    }
};

 Java:

/*
 * @Autor: SourDumplings
 * @Date: 2019-09-27 13:45:57
 * @Link: https://github.com/SourDumplings/
 * @Email: [email protected]
 * @Description: https://leetcode.com/problems/number-complement/
 */

class Solution
{
    public int findComplement(int num)
    {
        int s = 1;
        int e = 0;
        for (int i = 0; i < 31; i++)
        {
            if ((num & s) != 0)
            {
                e = i;
            }
            s <<= 1;
        }
        int mask = (1 << (e + 1)) - 1;
        return num ^ mask;
    }
}