leetcode 1309. Decrypt String from Alphabet to Integer Mapping
Given a string s
formed by digits ('0'
- '9'
) and '#'
. We want to map s
to English lowercase characters as follows:
- Characters (
'a'
to'i')
are represented by ('1'
to'9'
) respectively. - Characters (
'j'
to'z')
are represented by ('10#'
to'26#'
) respectively.
Return the string formed after mapping.
It's guaranteed that a unique mapping will always exist.
Example 1:
Input: s = "10#11#12" Output: "jkab" Explanation: "j" -> "10#" , "k" -> "11#" , "a" -> "1" , "b" -> "2".
Example 2:
Input: s = "1326#" Output: "acz"
Example 3:
Input: s = "25#" Output: "y"
Example 4:
Input: s = "12345678910#11#12#13#14#15#16#17#18#19#20#21#22#23#24#25#26#" Output: "abcdefghijklmnopqrstuvwxyz"
Constraints:
1 <= s.length <= 1000
s[i]
only contains digits letters ('0'
-'9'
) and'#'
letter.s
will be valid string such that mapping is always possible.
思路:判断一个数字字符应该转为a−j的字符还是应该和后一位数字字符结合转为j-z,应该看第三位字符,如果是#,则需要将其和后一位字符结合。
C++:
1 class Solution { 2 public: 3 string freqAlphabets(string s) { 4 string ans; 5 for (int i = 0; i < s.length(); ++i) { 6 int num = s[i] - '0'; 7 if (i + 2 < s.length() && (s[i + 2] == '#')) { 8 num = num * 10 + (s[i + 1] - '0'); 9 i += 2; 10 } 11 ans.push_back('a' + num - 1); 12 } 13 return ans; 14 } 15 };
python3:
1 import re 2 class Solution: 3 def freqAlphabets(self, s: str) -> str: 4 return ''.join([chr(int(i[:2]) + 96) for i in re.findall(r'\d\d#|\d', s)])
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