LeetCode 1578.替换所有的问号

原创已于 2022-04-02 16:40:28 修改 · 404 阅读

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于 2022-01-05 23:52:02 首次发布

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该博客主要讨论了LeetCode的一道题目，即如何在包含小写字母和问号的字符串中替换问号，使得最终字符串不包含连续重复字符。解决方案是遍历字符串，使用'a'、'b'、'c'替换问号，确保替换后的字符与相邻字符不相同。文章提供了C++的实现代码，并给出了两个示例解释了题目的要求和解题思路。

原题链接：LeetCode 1578.替换所有的问号

Given a string s containing only lowercase English letters and the ’?' character, convert all the ‘?’ characters into lowercase 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”.

Constraints:

1 <= s.length <= 100
s consist of lowercase English letters and ‘?’.

思路：
替换字符只要和他前后都不一样就可以，因此用a、b、c就足够
遍历a、b、c，和它前后都不一样的时候，替换即可

c++实现：

class Solution {
public:
    string modifyString(string s) {
        for(int i = 0; i < s.length(); i ++ ){
            if(s[i] == '?'){
                for(char c = 'a'; c <= 'c'; ++c){
                    if((i > 0 && s[i-1] == c) || (i < s.length()-1 && s[i+1] == c))
                        continue;
                    s[i] = c;
                    break;
                }   
            }
        }
        return s;
    }
};