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.
A:
class Solution { public: string freqAlphabets(string s) { vector<char> V; int n = s.length(); for(int i =0;i<n; ++i){ char val = 'a';// this value is never used though if(i+2<n && s[i+2] == '#'){ val = 'j' + stoi(s.substr(i,2)) - 10; i+=2; }else{ val = 'a' + (s[i] - '1'); } V.push_back(val); } string res(V.begin(), V.end()); return res; } };
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