Skip to content

Latest commit

 

History

History
148 lines (117 loc) · 3.95 KB

File metadata and controls

148 lines (117 loc) · 3.95 KB

中文文档

Description

Given a string word, return the sum of the number of vowels ('a', 'e', 'i', 'o', and 'u') in every substring of word.

A substring is a contiguous (non-empty) sequence of characters within a string.

Note: Due to the large constraints, the answer may not fit in a signed 32-bit integer. Please be careful during the calculations.

 

Example 1:

Input: word = "aba"
Output: 6
Explanation: 
All possible substrings are: "a", "ab", "aba", "b", "ba", and "a".
- "b" has 0 vowels in it
- "a", "ab", "ba", and "a" have 1 vowel each
- "aba" has 2 vowels in it
Hence, the total sum of vowels = 0 + 1 + 1 + 1 + 1 + 2 = 6. 

Example 2:

Input: word = "abc"
Output: 3
Explanation: 
All possible substrings are: "a", "ab", "abc", "b", "bc", and "c".
- "a", "ab", and "abc" have 1 vowel each
- "b", "bc", and "c" have 0 vowels each
Hence, the total sum of vowels = 1 + 1 + 1 + 0 + 0 + 0 = 3. 

Example 3:

Input: word = "ltcd"
Output: 0
Explanation: There are no vowels in any substring of "ltcd".

Example 4:

Input: word = "noosabasboosa"
Output: 237
Explanation: There are a total of 237 vowels in all the substrings.

 

Constraints:

  • 1 <= word.length <= 105
  • word consists of lowercase English letters.

Solutions

Python3

class Solution:
    def countVowels(self, word: str) -> int:
        ans, n = 0, len(word)
        for i, c in enumerate(word):
            if c in ['a', 'e', 'i', 'o', 'u']:
                ans += (i + 1) * (n - i)
        return ans

Java

class Solution {
    public long countVowels(String word) {
        long ans = 0;
        for (int i = 0, n = word.length(); i < n; ++i) {
            char c = word.charAt(i);
            if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') {
                ans += (long) (i + 1) * (n - i);
            }
        }
        return ans;
    }
}

TypeScript

function countVowels(word: string): number {
    const n = word.length;
    let ans = 0;
    for (let i = 0; i < n; i++) {
        let char = word.charAt(i);
        if (['a', 'e', 'i', 'o', 'u'].includes(char)) {
            ans += ((i + 1) * (n - i));
        }
    }
    return ans;
};

C++

class Solution {
public:
    long long countVowels(string word) {
        long long ans = 0;
        for (int i = 0, n = word.size(); i < n; ++i)
        {
            char c = word[i];
            if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') ans += (long long) (i + 1) * (n - i);
        }
        return ans;
    }
};

Go

func countVowels(word string) int64 {
	var ans int64
	n := len(word)
	for i, c := range word {
		if c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' {
			ans += int64((i + 1) * (n - i))
		}
	}
	return ans
}

...