跳转至

LeetCode: 482. License Key Formatting

1、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:

  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.

2、解题思路

​ 一开始没有理解题意,实际上,他的意思是,假如使用3个字符一组,那么所有的字符,除了第一个,其他的都是3个字符

char* licenseKeyFormatting(char* S, int K) {

    int length = strlen(S);

    int length_temp = length;

    int result_pos = length + length / K;

    char *temp = S + length - 1;

    char *result = (char *) calloc(length + length / K, sizeof(char));


    int count = K;
    while (length_temp-- > 0) {
        if (*temp != '-') {
            if (count == 0) {
                result[result_pos--] = '-';
                count = K;
            }
            if (*temp >= 'a' && *temp <= 'z') {
                result[result_pos--] = *temp - 32;
            } else {
                result[result_pos--] = *temp;
            }

            count--;
        }

        temp--;

    }

    return &result[result_pos+1];

}