LeetCode-1576. Replace all question marks

Here's a small alphabet and '?' String of characters s, Please put all the '?' Convert to a number of lowercase letters , So that the final string does not contain any Keep repeating The characters of .

Be careful ： you You can't Modify non '?' character .

Topic test cases guarantee except '?' character outside , There are no consecutive repeated characters .

At the completion of all transformations （ There may be no need to convert ） After the return of the final string . If there are multiple solutions , Please return to any of them . Can prove that , Under given constraints , The answer is always there .

Example 1：

Input ：s = "?zs"
Output ："azs"
explain ： This example has 25 Kind of solution , from "azs" To "yzs" All of them meet the requirements of the title . Only "z" It's an invalid modification , Because strings "zzs" There are two consecutive repetitions in 'z' .
Example 2：

Input ：s = "ubv?w"
Output ："ubvaw"
explain ： This example has 24 Kind of solution , Only replace it with "v" and "w" Does not meet the requirements of the topic . because "ubvvw" and "ubvww" All contain characters that repeat continuously .
Example 3：

Input ：s = "j?qg??b"
Output ："jaqgacb"
Example 4：

Input ：s = "??yw?ipkj?"
Output ："acywaipkja"
 

Tips ：

1 <= s.length <= 100

s Contains only lower case letters and '?' character

#include<iostream>
using namespace std;
class Solution {
public:
    string modifyString(string s) {
        if (s[0] == '?') {
            s[0] = getonechar(s[1]);
        }
        for (int i = 1; i < s.length() - 1; i++) {
            if (s[i] == '?') {
                s[i] = getonechar(s[i - 1], s[i + 1]);
            }
        }

        if (s[s.length() - 1] == '?') {
            s[s.length() - 1] = getonechar(s[s.length() - 2]);
        }

        return s;
    }

    char getonechar(char a) {
        for (char i = 'a'; i <= 'z'; i++) {
            if (i != a) {
                return i;
            }
        }
        return a;
    }
    char getonechar(char a, char b) {
        for (char i = 'a'; i <= 'z'; i++) {
            if (i != a && i != b) {
                return i;
            }
        }
        return a;
    }

};