LeetCode 482. License Key Formatting(密钥格式化)
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:
- 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.
解法一
思路
这题换一个说法可以是:将(忽略连接号的)字符串切分为n
份长度为K
的子串,其中第一份的长度可以少于K
;再将这些字串以连接号拼接为一个字符串。
所以这题的关键就是在第一份子串的处理上,这里的第一份就像是除法的余数部分。实际上,我们可以采用从后向前的遍历方法来简单高效地处理这个问题:原字符串从后向前遍历,(忽略原连接号)每遇到K
个字符便在其前面添加一个连接号,直到最后一份(实际上是第一份)不足K
个字符。这样是不是就像是模拟了一遍整数除法和求余的步骤呢?
当然这里还有一个细节问题要处理:当字符个数整除K
的时候,最后添加的那个连接号应该删去(多余)。
Python
class Solution:
def licenseKeyFormatting(self, S, K):
"""
:type S: str
:type K: int
:rtype: str
"""
result = ''
count_k = 0
for letter in reversed(S):
if letter != '-':
result += letter.upper()
count_k += 1
if count_k == K:
result += '-'
count_k = 0
if len(result) != 0 and result[-1] == '-':
result = result[:-1]
return result[::-1]
Java
class Solution {
public:
string licenseKeyFormatting(string S, int K) {
string result = "";
int countK = 0;
for (int i = S.length() - 1; i >= 0; i--) {
char letter = S[i];
if (letter != '-') {
result += toupper(letter);
countK++;
if (countK == K) {
result += '-';
countK = 0;
}
}
}
if (result.length() != 0 && result[result.length() - 1] == '-') {
result.pop_back();
}
reverse(result.begin(), result.end());
return result;
}
};
C++
public class Solution {
public String licenseKeyFormatting(String S, int K) {
StringBuilder result = new StringBuilder();
int countK = 0;
for (int i = S.length() - 1; i >= 0; i--) {
char letter = S.charAt(i);
if (letter != '-') {
result.append(letter);
countK++;
if (countK == K) {
result.append('-');
countK = 0;
}
}
}
if (result.length() != 0 && result.charAt(result.length() - 1) == '-') {
result.deleteCharAt(result.length() - 1);
}
return result.reverse().toString().toUpperCase();
}
}
java 和C++代码写反了吧。。。