LeetCode解题报告（339) -- 1704. Determine if String Halves Are Alike

Problem

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:

1
2
3

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:

1 2	Input: s = "MerryChristmas" Output: false

Example 4:

1 2	Input: s = "AbCdEfGh" Output: true

Constraints:

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

Analysis

题目给出一个字符串，从中间对半切开，要求我们判断前半部分和后半部分中包含的元音字母个数是否相同。

因为是前后两半对比，所以遍历一半就可以了，每个字符都判断下是否元音，前后两部分各自统计一下元音出现的次数，最后看看两者是否相等即可。

Solution

注意大小写的问题，

Code

class Solution {
public:
    bool halvesAreAlike(string s) {
        int size = s.size();
        int num1 = 0, num2 = 0;
        for (int i = 0; i < size / 2; i++) {
            if (isVowel(s[i])) {
                num1++;
            }
            if (isVowel(s[size - 1 - i])) {
                num2++;
            }
        }
        // cout << num1 << " " << num2 << endl;
        return num1 == num2;
    }
private:
    bool isVowel(char c) {
        if (c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U' ||
           c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') {
            return true;
        }
        return false;
    }
};

Summary

这是一道非常简单的字符串遍历题目，没什么难点。这道题目的分享到这里，感谢你的支持！