Merge branch 'master' into 'master'

feat: 简单的一个字符中元音字母数量算法

See merge request csdn/contest_exercises_template!1

exercises/liuwei/简单/一个字符串作为输入返回该字符串中元音字母的数量/exercies.md

+# 一个字符串作为输入返回该字符串中元音字母的数量
+
+请编写一个函数 countVowels，该函数接受一个字符串作为输入，然后返回该字符串中元音字母的数量。
+
+## 输入描述
+一个字符串。
+ 
+## 输出描述
+
+ 一个数字，表示该字符串中元音字母的数量。
+
+## 输入样例
+console.log(countVowels('hello')); // 输出 2
+ 
+
+## 输出样例
+console.log(countVowels('javascript')); // 输出 3
+console.log(countVowels('world')); // 输出 1
+console.log(countVowels('')); // 输出 0
+console.log(countVowels('aeiou')); // 输出 5
+console.log(countVowels('AEIOU')); // 输出 5
+console.log(countVowels('hjklmnpqrstvwxyz')); // 输出 0
+console.log(countVowels('Hello World!')); // 输出 3
+console.log(countVowels('To be or not to be, that is the question.')); // 输出13
+console.log(countVowels('Brevity is the soul of wit.')); // 输出 8
+
+## 提示
+遍历字符串，依次检查每个字符是否是元音字母；
+如果是，则将计数器加1；
+最后返回计数器的值。

exercises/liuwei/简单/一个字符串作为输入返回该字符串中元音字母的数量/solution.js

+function countVowels(str) {
+    const vowels = ['a', 'e', 'i', 'o', 'u'];
+    let count = 0;
+    for (let i = 0; i < str.length; i++) {
+      if (vowels.indexOf(str[i].toLowerCase()) !== -1) {
+        count++;
+      }
+    }
+    return count;
+  }
+  
\ No newline at end of file

exercises/liuwei/简单/一个字符串作为输入返回该字符串中元音字母的数量/test_cases/10.in

+19
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exercises/liuwei/简单/一个字符串作为输入返回该字符串中元音字母的数量/test_cases/8.in

+79
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exercises/liuwei/简单/一个字符串作为输入返回该字符串中元音字母的数量/test_cases/9.in

+38
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