描述
You are given a 0-indexed string s that has lowercase English letters in its even indices and digits in its odd indices.
There is a function shift(c, x), where c is a character and x is a digit, that returns the xth character after c.
- For example, shift('a', 5) = 'f' and shift('x', 0) = 'x'.
For every odd index i, you want to replace the digit s[i] with shift(s[i-1], s[i]).
Return s after replacing all digits. It is guaranteed that shift(s[i-1], s[i]) will never exceed 'z'.
Example 1:
Input: s = "a1c1e1"
Output: "abcdef"
Explanation: The digits are replaced as follows:
- s[1] -> shift('a',1) = 'b'
- s[3] -> shift('c',1) = 'd'
- s[5] -> shift('e',1) = 'f'
Example 2:
Input: s = "a1b2c3d4e"
Output: "abbdcfdhe"
Explanation: The digits are replaced as follows:
- s[1] -> shift('a',1) = 'b'
- s[3] -> shift('b',2) = 'd'
- s[5] -> shift('c',3) = 'f'
- s[7] -> shift('d',4) = 'h'
Note:
1 <= s.length <= 100
s consists only of lowercase English letters and digits.
shift(s[i-1], s[i]) <= 'z' for all odd indices i.
解析
根据题意，就是将字符串 s 中的索引为奇数时候的数字，变为前一个字符按照英文字母顺序的后面第 i 个字母，思路很简单，直接遍历字符串 s ，然后将需要变化的字母进行转换即可得到答案。
解答
class Solution(object):
def replaceDigits(self, s):
"""
:type s: str
:rtype: str
"""
r = ""
for i in range(len(s)):
if i % 2 == 0:
r += s[i]
else:
r += chr(ord(s[i - 1]) + int(s[i]))
return r
运行结果
Runtime: 24 ms, faster than 41.12% of Python online submissions for Replace All Digits with Characters.
Memory Usage: 13.5 MB, less than 44.16% of Python online submissions for Replace All Digits with Characters.
原题链接：leetcode.com/problems/re…
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