You are given a string s of even length. Split this string into two halves of equal lengths, and let a be the first half and b be the second half.

Two strings are alike if they have the same number of vowels ('a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'). Notice that s contains uppercase and lowercase letters.

Return true if a and b are alike. Otherwise, return false.

Example 1:

Input: s = "book" Output: true Explanation: a = "bo" and b = "ok". a has 1 vowel and b has 1 vowel. Therefore, they are alike. 

Example 2:

Input: s = "textbook" Output: false Explanation: a = "text" and b = "book". a has 1 vowel whereas b has 2. Therefore, they are not alike. Notice that the vowel o is counted twice. 

Example 3:

Input: s = "MerryChristmas" Output: false 

Example 4:

Input: s = "AbCdEfGh" Output: true 

Constraints:

• 2 <= s.length <= 1000
• s.length is even.
• s consists of uppercase and lowercase letters.

给你一个偶数长度的字符串 s 。将其拆分成长度相同的两半，前一半为 a ，后一半为 b 。两个字符串 相似 的前提是它们都含有相同数目的元音（'a'，'e'，'i'，'o'，'u'，'A'，'E'，'I'，'O'，'U'）。注意，s 可能同时含有大写和小写字母。如果 a 和 b 相似，返回 true ；否则，返回 false 。

• 简单题。依题意，分别统计前半段元音字母的个数和后半段元音字母的个数，个数相同则输出 true，不同就输出 false。

package leetcode funchalvesAreAlike(sstring) bool { returnnumVowels(s[len(s)/2:]) ==numVowels(s[:len(s)/2]) } funcnumVowels(xstring) int { res:=0for_, c:=rangex { switchc { case'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U': res++ } } returnres }