题解 | 凯撒解密

#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 进行凯撒解密
# @param password string字符串 旺仔哥哥的密码
# @param n int整型 每个字符加密过程中错位的次数
# @return string字符串
#
class Solution:
    def decodeWangzai(self , password: str, n: int) -> str:
        # write code here
        s = ''
        for i in password:
            s += chr(ord('a') + (ord(i) - n - ord('a')) % 26)
        return s
"""
a % b = a - b * floor(a / b)
对字符'a'来说，
（ord('a') - 1 - ord('a')) % 26 == (-1) % 26 == -1 - 26 * floor(-1/26) == -1 -26 * (-1) == 25
"""