[LeetCode] 482. License Key Formatting 注册码格式化

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.

给一个字符串的注册码S，按要求对注册码进行格式化，每四个字符后面跟一个短横杠，每一部分的长度为K，第一部分长度可以小于K，所有字母必须是大写的。

解法：用数组记录字符，从后往前循环字符串，当前字符不是短横杠，就变大写加入数组，每隔四个字母加一个短横杠，最后在转换成字符串。

最重要的是判断4个字符的方法： if len(res) % (K + 1) == K

Java:

?

1 2 3 4 5 6 7 8 9 10 11

class Solution {     public String licenseKeyFormatting(String S, int K) {         StringBuilder sb = new StringBuilder();         for (int i = S.length() - 1; i >= 0; i--) {             if (S.charAt(i) != '-') {                 sb.append(sb.length() % (K + 1) == K ? "-" : "").append(S.charAt(i));             }         }         return sb.reverse().toString().toUpperCase();     } }

Java:

?

1 2 3 4 5 6 7

public String licenseKeyFormatting(String s, int k) {     StringBuilder sb = new StringBuilder();     for (int i = s.length() - 1; i >= 0; i--)         if (s.charAt(i) != '-')             sb.append(sb.length() % (k + 1) == k ? '-' : "").append(s.charAt(i));     return sb.reverse().toString().toUpperCase(); }

Python: wo

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

class Solution(object):     def licenseKeyFormatting(self, S, K):         """         :type S: str         :type K: int         :rtype: str         """         res = []         for i in reversed(range(len(S))):             if S[i] != '-':                 if len(res) % (K + 1) == K:                     res.append('-')                 res.append(S[i].upper())                           return ''.join(s for s in reversed(res))

Python: wo

?

1 2 3 4 5 6 7 8 9 10

class Solution(object):     def licenseKeyFormatting(self, S, K):         res = []         for c in reversed(S):             if c != '-':                 if len(res) % (K + 1) == K:                     res.append('-')                 res.append(c.upper())                           return ''.join(r for r in reversed(res))

Python:

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

class Solution(object):     def licenseKeyFormatting(self, S, K):         """         :type S: str         :type K: int         :rtype: str         """         S = S.upper().replace('-','')         size = len(S)         s1 = K if size%K==0 else size%K         res = S[:s1]         while s1<size:             res += '-'+S[s1:s1+K]             s1 += K         return res

Python:

?

1 2 3 4 5 6 7 8 9

class Solution:     def licenseKeyFormatting(self, S, K):         """         :type S: str         :type K: int         :rtype: str         """         S = S.replace("-", "").upper()[::-1]         return '-'.join(S[i:i+K] for i in range(0, len(S), K))[::-1]

Python:

?

1 2 3 4 5 6 7 8 9 10 11 12

class Solution:     def licenseKeyFormatting(self, S, K):         """         :type S: str         :type K: int         :rtype: str         """         def chunks(l, n):             for i in range(0, len(l), n):                 yield l[i:i+n]         s = S[::-1].upper().replace('-', '')           return '-'.join(list(chunks(s, K)))[::-1]

C++:

?

1 2 3 4 5 6 7 8 9 10 11 12

class Solution { public:     string licenseKeyFormatting(string S, int K) {         string res = "";         for (int i = (int)S.size() - 1; i >= 0; --i) {             if (S[i] != '-') {                 ((res.size() % (K + 1) - K) ? res : res += '-') += toupper(S[i]);             }         }         return string(res.rbegin(), res.rend());     } };

　　

 

 

All LeetCode Questions List 题目汇总