闯关leetcode——3136. Valid Word

2025/1/5 23:33:20 来源：https://blog.csdn.net/breaksoftware/article/details/142669228 浏览: 次关键词：闯关leetcode——3136. Valid Word

大纲

题目
- 地址
- 内容
解题
- 代码地址

题目

地址

https://leetcode.com/problems/valid-word/description/

内容

A word is considered valid if:

It contains a minimum of 3 characters.
It contains only digits (0-9), and English letters (uppercase and lowercase).
It includes at least one vowel.
It includes at least one consonant.

You are given a string word.

Return true if word is valid, otherwise, return false.

Notes:

‘a’, ‘e’, ‘i’, ‘o’, ‘u’, and their uppercases are vowels.
A consonant is an English letter that is not a vowel.

Example 1:

Input: word = “234Adas”
Output: true
Explanation:
This word satisfies the conditions.

Example 2:

Input: word = “b3”
Output: false
Explanation:
The length of this word is fewer than 3, and does not have a vowel.

Example 3:

Input: word = "a3 $e " O u tp u t : f a l se E x pl ana t i o n : T hi s w or d co n t ain s a^{'}$ ’ character and does not have a consonant.

Constraints:

1 <= word.length <= 20
word consists of English uppercase and lowercase letters, digits, ‘@’, ‘#’, and ‘$’.

解题

这题要检测一个数是否符合以下特点：

至少包含3个字符
只能包含大小写字母和数字
至少有一个a、e、i、o、u大写或小写的字母
至少包含一个非a、e、i、o、u大写或小写的字母

解法也很简单，按这些条件把规则写好即可。

#include <string>
using namespace std;class Solution {
public:bool isValid(string word) {if (word.size() < 3) {return false;}bool vowel = false;bool consonant = false;for (char c : word) {if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u'|| c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U') {vowel = true;} else if ( (c >= 'a' && c <= 'z')|| (c >= 'A' && c <= 'Z')){consonant = true;} else if (c >= '0' && c <= '9') {continue;}else {return false;}}return vowel && consonant;}
};