# 问题 原题链接：[https://leetcode.com/problems/license-key-formatting/](https://leetcode.com/problems/license-key-formatting/) 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. # 分析过程 - 输入：只有数字、字母和横杠的字符串以及一个分段长度 k - 输出：按 k 分段格式化以后的字符串 - 思路：由于输出的长度有可能超过输入，因此不能在原字符串修改，需要开辟临时空间。由于题目规定只有第一段长度可以小于 k，因此采取反向输入的方法，在临时空间中反向输入字符。输入时记录长度，如果到达 k 就分段即可。 # 解决方法 ```cpp class Solution { public: string licenseKeyFormatting(string S, int K) { // 题目已规定输入是合法的 const char *strPtr = S.c_str(); int len = (int)strlen(strPtr); // 分配临时空间 // 题目规定输入最长12000，如果k是1，结果最长24000 // 但是testcase里有一个长度四万的case（- -|||） char buff[80001]; memset(buff, 0, 80001); // 从后往前输入，定义两个指针，分别指向源字符串的末尾和buffer的末尾 const char *srcPtr = strPtr + len - 1; char *dstPtr = buff + 80001 - 2; // 字符分段计数，用于和k比较 int kCounter = 0; // 循环结束条件是源字符串扫描结束或buff用完 while (srcPtr >= strPtr && dstPtr >= buff) { if (kCounter >= K) { // 如果已经达到k，则输出-并清空计数 *dstPtr = '-'; kCounter = 0; --dstPtr; } else { // 如果没有达到计数，则输入字符 // 首先跳过所有- while (srcPtr >= strPtr && dstPtr >= buff && *srcPtr == '-') { --srcPtr; } // 检查是不是到循环结束条件了 if (srcPtr < strPtr || dstPtr < buff) { break; } // 输入字符并增加计数和移动指针 char c = *srcPtr; *dstPtr = c >= 'a' && c <= 'z' ? c - 32 : c; ++kCounter; --srcPtr; --dstPtr; } } // 跳过开头的- while (*(dstPtr + 1) == '-' ) { dstPtr++; } // 返回结果 return string(dstPtr + 1); } }; ```