LeetCode Maximum Number of Vowels in a Substring of Given Length

5417. Maximum Number of Vowels in a Substring of Given Length

Given a string s and an integer k.

Return the maximum number of vowel letters in any substring of s with length k.

Vowel letters in English are (a, e, i, o, u).

Example 1:

Input: s = "abciiidef", k = 3
Output: 3
Explanation: The substring "iii" contains 3 vowel letters.

Example 2:

Input: s = "aeiou", k = 2
Output: 2
Explanation: Any substring of length 2 contains 2 vowels.

Example 3:

Input: s = "leetcode", k = 3
Output: 2
Explanation: "lee", "eet" and "ode" contain 2 vowels.

Example 4:

Input: s = "rhythms", k = 4
Output: 0
Explanation: We can see that s doesn't have any vowel letters.

Example 5:

Input: s = "tryhard", k = 4
Output: 1

Constraints:

  • 1 <= s.length <= 10^5
  • s consists of lowercase English letters.
  • 1 <= k <= s.length

给定字符串s和长度k,问s中长度为k的子串中,包含最多元音字母的个数。

简单题,直接滑动窗口+双指针计数即可:

class Solution {
public:
	bool IsVol(const char &c) {
		return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
	}
	int maxVowels(string s, int k) {
		int ans = 0;
		int i = 0, j = k - 1, n = s.size();
		for (int k = i; k <= j; ++k) {
			if (IsVol(s[k]))++ans;
		}
		int cur = ans;
		while (j < n - 1) {
			++j;
			if (IsVol(s[j]))++cur;
			if (IsVol(s[i]))--cur;
			++i;
			ans = max(ans, cur);
		}
		return ans;
	}
};

本代码提交AC。

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