题解 | #HJ21 简单密码#

解法1-条件判断

这题思路非常清晰，最直接的方式就是通过条件判断来转换每个字符。

def char_to_num(char):
    if char == 'Z':
        return 'a'
    if char.isupper():
        char = chr(ord(char.lower()) + 1)
        return char
    if char in 'abc': return '2'
    elif char in 'def': return '3'
    elif char in 'ghi': return '4'
    elif char in 'jkl': return '5'
    elif char in 'mno': return '6'
    elif char in 'pqrs': return '7'
    elif char in 'tuv': return '8'
    elif char in 'wxyz': return '9'
    else: return char

line = input()
s = "".join(char_to_num(char) for char in line)
print(s)

解法2-字典映射

看到A, B这两个整齐对应的序列，第一时间就想到了zip，分分钟帮你映射成字典，再结合列表生成式，更简洁了！

A = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ'
B = '22233344455566677778889999bcdefghijklmnopqrstuvwxyza'
mapping = dict(zip(A, B))
s = ''.join([mapping[c] if c in A else c for c in input()])
print(s)