【leetcode】1576. Replace All ?'s to Avoid Consecutive Repeating Characters
题目如下:
Given a string
s
containing only lower case English letters and the '?' character, convert all the '?' characters into lower case letters such that the final string does not contain any consecutive repeating characters. You cannot modify the non '?' characters.It is guaranteed that there are no consecutive repeating characters in the given string except for '?'.
Return the final string after all the conversions (possibly zero) have been made. If there is more than one solution, return any of them. It can be shown that an answer is always possible with the given constraints.
Example 1:
Input: s = "?zs" Output: "azs" Explanation: There are 25 solutions for this problem. From "azs" to "yzs", all are valid. Only "z" is an invalid modification as the string will consist of consecutive repeating
characters in "zzs".Example 2:
Input: s = "ubv?w" Output: "ubvaw" Explanation: There are 24 solutions for this problem. Only "v" and "w" are invalid modifications as the strings will consist of consecutive
repeating characters in "ubvvw" and "ubvww".Example 3:
Input: s = "j?qg??b" Output: "jaqgacb"Example 4:
Input: s = "??yw?ipkj?" Output: "acywaipkja"Constraints:
1 <= s.length <= 100
s
contains only lower case English letters and'?'
.
解题思路:题目不难,只要把?替换成与前面和后面的字符都不一致的字符即可,注意要考虑?在字符首尾出现的情况。
代码如下:
class Solution(object): def modifyString(self, s): """ :type s: str :rtype: str """ res = '' char_list = [chr(i) for i in range(97,123)] for i in range(len(s)): if s[i] != '?': res += s[i] continue elif len(s) == 1: res += 'a' elif i == len(s) - 1: for char in char_list: if char != res[-1]: res += char break elif i == 0: for char in char_list: if char != s[i+1]: res += char break else: for char in char_list: if char != s[i+1] and char != res[-1]: res += char break return res