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.
A:
class Solution { public: string licenseKeyFormatting(string S, int K) { int count = 0; // count of chars in current group vector<char> V; for(int i= S.length()-1;i>=0;--i) { char ch = S[i]; if(ch == '-') continue; V.push_back(toupper(ch)); ++count; if(count == K) { V.push_back('-'); count =0; } } if(not V.empty() && V.back() =='-') //Calling vector.back() on an empty container causes undefined behavior. V.pop_back(); string res(V.rbegin(), V.rend()); return res; } };
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