1309 Decrypt String from Alphabet to Integer Mapping 解码字母到整数映射
Description:
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:
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.
题目描述:
给你一个字符串 s，它由数字（'0' - '9'）和 '#' 组成。我们希望按下述规则将 s 映射为一些小写英文字符：
字符（'a' - 'i'）分别用（'1' - '9'）表示。
字符（'j' - 'z'）分别用（'10#' - '26#'）表示。
返回映射之后形成的新字符串。
题目数据保证映射始终唯一。
示例 :
示例 1：
输入：s = "10#11#12"
输出："jkab"
解释："j" -> "10#" , "k" -> "11#" , "a" -> "1" , "b" -> "2".
示例 2：
输入：s = "1326#"
输出："acz"
示例 3：
输入：s = "25#"
输出："y"
示例 4：
输入：s = "12345678910#11#12#13#14#15#16#17#18#19#20#21#22#23#24#25#26#"
输出："abcdefghijklmnopqrstuvwxyz"
提示：
1 <= s.length <= 1000
s[i] 只包含数字（'0'-'9'）和 '#' 字符。
s 是映射始终存在的有效字符串。
思路:
从字符串末尾开始遍历, 要么找到 ‘#’找前两位, 要么单独的 1位转换成字母, 插到结果的开头即为答案
时间复杂度O(n), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
string freqAlphabets(string s)
{
string result;
for (int i = s.size() - 1; i > -1; i--) result.insert(0, 1, 'a' + (s[i] == '#' ? s[--i] - '1' + 10 * (s[--i] - '0') : s[i] - '1'));
return result;
}
};
Java:
class Solution {
public String freqAlphabets(String s) {
StringBuilder sb = new StringBuilder();
for (int i = s.length() - 1; i > -1; i--) sb.insert(0, (char)('a' + (s.charAt(i) == '#' ? s.charAt(--i) - '1' + 10 * (s.charAt(--i) - '0') : s.charAt(i) - '1')));
return sb.toString();
}
}
Python:
class Solution:
def freqAlphabets(self, s: str) -> str:
return re.sub(r'\d\d#|\d', lambda x: chr(int(x.group()[:2]) + 96), s)
