【Day 43 】2023-03-28 - 1456. 定长子串中元音的最大数目

[azl397985856]

1456. 定长子串中元音的最大数目

入选理由

暂无

题目地址

https://leetcode-cn.com/problems/maximum-number-of-vowels-in-a-substring-of-given-length

前置知识

暂无

题目描述

给你字符串 s 和整数 k 。

请返回字符串 s 中长度为 k 的单个子字符串中可能包含的最大元音字母数。

英文中的 元音字母 为（a, e, i, o, u）。

示例 1：

输入：s = "abciiidef", k = 3
输出：3
解释：子字符串 "iii" 包含 3 个元音字母。
示例 2：

输入：s = "aeiou", k = 2
输出：2
解释：任意长度为 2 的子字符串都包含 2 个元音字母。
示例 3：

输入：s = "leetcode", k = 3
输出：2
解释："lee"、"eet" 和 "ode" 都包含 2 个元音字母。
示例 4：

输入：s = "rhythms", k = 4
输出：0
解释：字符串 s 中不含任何元音字母。
示例 5：

输入：s = "tryhard", k = 4
输出：1

提示：

1 <= s.length <= 10^5
s 由小写英文字母组成
1 <= k <= s.length

[DrinkMorekaik]

var maxVowels = function(s, k) {
  const letters = ['a', 'o', 'e', 'i', 'u']
  let count = 0
  for(let i = 0; i < k; i++) {
    letters.includes(s[i]) && count++
  }
  let current = count
  for (let j = k; j < s.length; j++) {
    letters.includes(s[j]) && current++
    letters.includes(s[j - k]) && current--
    count = Math.max(count, current)
  }
  return count
};

[JiangyanLiNEU]

• Sliding Window
• T = O(N)

• S = O(N)

  class Solution:
  def maxVowels(self, s: str, k: int) -> int:
      vowels = set(['a', 'e', 'i', 'o', 'u'])
      isVowel = [1 if char in vowels else 0 for char in s]
  
      accumulate = [0]+[isVowel[0]]
      for i in range(1, len(s)):
          accumulate.append(isVowel[i]+accumulate[-1])
  
      res = 0
      for i in range(k, len(s)+1):
          res = max(res, accumulate[i]-accumulate[i-k])
      return res

[JasonQiu]

class Solution:
    def maxVowels(self, s: str, k: int) -> int:
        current = 0
        result = 0
        for pos in range(k):
            if s[pos] in "aeiou":
                current += 1
        result = current
        for pos in range(1, len(s) - k + 1):
            if s[pos - 1] in "aeiou":
                current -= 1
            if s[pos + k - 1] in "aeiou":
                current += 1
            result = max(result, current)
        return result

Time: O(n) Space: O(1)

[bingzxy]

思路

滑动窗口

代码

class Solution {
    public int maxVowels(String s, int k) {
        int ans = 0, len = s.length();
        Set<Character> set = new HashSet<> () {{
            add('a');
            add('e');
            add('i');
            add('o');
            add('u');
        }};
        for (int i = 0; i < k; i++) {
            if (set.contains(s.charAt(i))) {
                ans++;
            }
        }

        int left = 0, right = k - 1;
        int cur = ans;
        while (right < len) {
            int tmp = cur;
            right = right + 1;
            if (right >= len) {
                return ans;
            }
            if (set.contains(s.charAt(right))) {
                tmp++;
            }
            if (set.contains(s.charAt(left))) {
                tmp--;
            }
            left++;
            cur = tmp;
            ans = Math.max(ans, tmp);
        }
        return ans;
    }
}

复杂度分析

• 时间复杂度：O(n)
• 空间复杂度：O(1)

[xb798298436]

var maxVowels = function (s, k) { const dict = new Set(["a", "e", "i", "o", "u"]); let ret = 0; for (let i = 0; i < k; i++) { if (dict.has(s[i])) ret++; }

let temp = ret; for (let i = k, j = 0; i < s.length; i++, j++) { if (dict.has(s[i])) temp++; if (dict.has(s[j])) temp--;

ret = Math.max(temp, ret);

}

return ret; };

[kofzhang]

思路

滑动窗口，双指针

复杂度

时间复杂度：O(n) 空间复杂度：O(1)

代码

class Solution:
    def maxVowels(self, s: str, k: int) -> int:
        curmax = 0
        cur = 0
        left = 0
        for index,c in enumerate(s):
            if c in "aeiou":
                cur += 1
            if index-left+1>k:
                if s[left] in "aeiou":
                    cur -=1
                left += 1            
            curmax = max(curmax,cur)
        return curmax

[bookyue]

TC: O(n)
SC: O(1)

    public int maxVowels(String s, int k) {
        int cnt = 0;
        for (int i = 0; i < k; i++) {
            if (isVowel(s.charAt(i)))
                cnt++;
        }

        int ans = cnt;
        for (int i = k; i < s.length(); i++) {
            if (isVowel(s.charAt(i - k)))
                cnt--;

            if (isVowel(s.charAt(i)))
                cnt++;

            ans = Math.max(ans, cnt);
            if (ans == k) return k;
        }

        return ans;
    }

    private boolean isVowel(char c) {
        return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
    }

[NorthSeacoder]

/**
 * @param {string} s
 * @param {number} k
 * @return {number}
 */
var maxVowels = function(s, k) {
    const vowels = 'aeiou';
    let count = 0;
    let res = 0;
    const len = s.length;
    for (let i = 0; i < Math.min(k, len); i++) {
        if (vowels.includes(s[i])) res++;
    }
    if (k >= len) return res;
    let left = 0;
    let right = k;
    count = res;
    while (right <= len) {
        if (vowels.includes(s[right])) count++;
        if (vowels.includes(s[left])) count--;
        res = Math.max(res, count);
        right++;
        left++;
    }
    return res;
};

• TC:O(n);
• SC:O(1)

[csthaha]

/**
 * @param {string} s
 * @param {number} k
 * @return {number}
 */
var maxVowels = function(s, k) {
    //固定长度的滑动窗口。
    let slow = 0;

    let max = 0
    let ans = 0;
    const vowel = ['a', 'e','i', 'o', 'u']
    for(let i = 0; i < s.length; i++) {
        if(i - slow + 1 > k) {
            slow++
            if(vowel.includes(s[slow - 1])) {
                max--;
            }
        }
        if(vowel.includes(s[i])) {
            max++;
        }
        ans = Math.max(ans, max)
    }
    return ans;
};

[Fuku-L]

代码

public class Q1456MaximumNumberOfVowelsInASubstringOfGivenLength {
    public static void main(String[] args) {
        Solution solution = new Q1456MaximumNumberOfVowelsInASubstringOfGivenLength().new Solution();
        System.out.println(solution.maxVowels("weallloveyou", 7));
    }

    class Solution {
        public int maxVowels(String s, int k) {
           String vowels = "aeiou";
           int n = s.length();
            int l = 0, r = (l + k) < n ? l + k : n;
            int cnt = 0;
            for (int i = l; i < r; i++) {
                if (vowels.indexOf(s.charAt(i)) >= 0){
                    cnt ++;
                }
            }
            int ans = cnt;
            for(int i = r; i < n; i++){
                if (vowels.indexOf(s.charAt(i - k)) >= 0){
                    cnt--;
                }
                if (vowels.indexOf(s.charAt(i)) >= 0){
                    cnt++;
                }
                ans = ans > cnt ? ans : cnt;
            }
            return ans;
        }
    }
}

复杂度分析

• 时间复杂度：O(|n|)，其中 n 为数组长度。
• 空间复杂度：O(1)

[wangqianqian202301]

思路

一个循环

代码

class Solution:
    def maxVowels(self, s: str, k: int) -> int:
        def isVowel(c: str) -> bool:
            return c == 'a' or c == 'e' or c == 'i' or c == 'o' or c == 'u'

        length = len(s)
        max = float("-inf")
        count = 0
        for i, c in enumerate(s):
            if i < k:
                if isVowel(c):
                    count = count + 1
            else:
                if isVowel(c):
                    count = count + 1
                if isVowel(s[i - k: i - k + 1]):
                    count = count - 1
            max = max if max > count else count
        return max

复杂度

O(n)

[linlizzz]

思路

维护一个长度为k的队列，记录里面的元音字母数res，依次沿着字符串s往后移动，移除或加入一个字母都判断一下是否是元音字母,对应res加减

代码

import collections
class Solution:
    def maxVowels(self, s: str, k: int) -> int:
        vowel = ['a', 'e', 'i', 'o', 'u']
        res, ans = 0, 0
        sub_s = collections.deque()
        for i in s:
            if len(sub_s) == k:
                l = sub_s.popleft()
                if l in vowel:
                    res -= 1
            sub_s.append(i)
            if i in vowel:
                res += 1
            if res == k:
                return k
            ans = max(ans, res)
        return ans

复杂度分析

T(n) = O(n)

S(n) = O(1)

[Abby-xu]

class Solution:
    def maxVowels(self, s: str, k: int) -> int:
        i,c,m,n=0,0,0,len(s)
        for j in range(n):
            if s[j] in 'aeiou':
                c+=1
            if j-i+1==k:
                m=max(m,c)
                if s[i] in 'aeiou':
                    c-=1
                i+=1
        return m

[FireHaoSky]

思路：滑动窗口

代码：python

class Solution:
    def maxVowels(self, s: str, k: int) -> int:
        def isVoxel(ch):
            return int(ch in 'aeiou')

        n = len(s)
        vowel_count = sum(1 for i in range(k) if isVoxel(s[i]))
        ans = vowel_count
        for i in range(k,n):
            vowel_count += isVoxel(s[i]) - isVoxel(s[i-k])
            ans = max(ans, vowel_count)
        return ans

复杂度分析：

"""
时间复杂度：o(n)
空间复杂度：o(1)
"""

[Jetery]

class Solution {
public:

    bool check(char c) {
        return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
    }

    int maxVowels(string s, int k) {
        int cur = 0, ans = 0;
        queue<char> q;
        for (char c : s) {
            q.push(c);
            if (q.size() > k) {
                char del = q.front();
                q.pop();
                if (check(del)) cur--;
            }
            if (check(c)) cur++;
            ans = max(ans, cur);
        }
        return ans;
    }
};

[jmaStella]

思路

滑动窗口 end每一个loop开始向右滑动一次，记录currCount start滑动条件： end-start>k，超过窗口大小start向右滑动，updatecurrCount

代码

public int maxVowels(String s, int k) {
        int start =0;
        int end = 0;
        Set<Character> vowel = new HashSet<>(Arrays.asList('a','e','i','o','u'));
        int result =0;
        int currCount = 0;

        while(end < s.length()){
            if(vowel.contains(s.charAt(end))){
                currCount++;
            }
            end++;

            while(end - start >k){
                if(vowel.contains(s.charAt(start))){
                    currCount--;
                }
                start++;
            }
            result = Math.max(result, currCount);
        }
        return result;
    }

复杂度

时间 O(N) 空间 O(`)

[wangzh0114]

思路

• 滑动窗口

  代码

  class Solution {
  public:
  int isVowel (char c){
      if(c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u')
          return 1;
      return 0;
  }
  int maxVowels(string s, int k) {
      int cnt = 0, ans = 0;
      for(int i = 0; i < s.size(); i++){
          cnt += isVowel(s[i]);
          if(i >= k){
              cnt -= isVowel(s[i - k]);
          }
          ans = max(ans, cnt);
      }
      return ans;
  }
  };

  复杂度分析

• TC:O(n)
• SC:O(1)

[huifeng248]

class Solution:
    def maxVowels(self, s: str, k: int) -> int:
        s_list = list(s)
        vowels = ['a', 'e', 'i', 'o', 'u']
        cnt = 0
        for i in range(k):
            if s_list[i] in vowels:
                cnt = cnt + 1
        max_num = cnt
        for end in range(k,len(s)):
            start = end - k + 1
            if s_list[start-1] in vowels:
                cnt = cnt - 1
            if s_list[end] in vowels:
                cnt = cnt + 1
            max_num = max(max_num, cnt)

        return max_num  

[Bochengwan]

思路

滑动窗口

代码

class Solution {
    public int maxVowels(String s, int k) {
        int n = s.length();
        int vowel_count = 0;
        for (int i = 0; i < k; ++i) {
            vowel_count += isVowel(s.charAt(i));
        }
        int ans = vowel_count;
        for (int i = k; i < n; ++i) {
            vowel_count += isVowel(s.charAt(i)) - isVowel(s.charAt(i - k));
            ans = Math.max(ans, vowel_count);
        }
        return ans;
    }

    public int isVowel(char ch) {
        return ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u' ? 1 : 0;
    }
}

复杂度分析

• 时间复杂度：O(len(s))
• 空间复杂度：O(1)

[harperz24]

class Solution:
    def maxVowels(self, s: str, k: int) -> int:
        vow = set(['a', 'e', 'i', 'o', 'u'])

        # construct first window
        count = 0
        for ch in s[:k]:
           count += ch in vow

        # slide from k to len(s) - 1
        # update max_count dynamically
        left = 0
        max_count = count    
        for i in range(k, len(s)):
            count -= s[left] in vow
            count += s[i] in vow
            left += 1
            max_count = max(max_count, count)
        return max_count

        # sliding window
        # time: O(n)
        # space: O(5)

[duke-github]

思路

滑动窗口 循环s中的字符串 下一个为元音则+1 当遍历的长度大于k以后 k之前的每个元音长度-1 取最大值

复杂度

时间复杂度O(n) 空间复杂度O(n)

代码

 public static int maxVowels(String s, int k) {
        char[] sChar = s.toCharArray();
        boolean[] vowels = new boolean[123];
        vowels['a'] = vowels['e'] = vowels['i'] = vowels['o'] = vowels['u'] = true;
        int vowelLen = 0;
        int ans = 0;
        for (int i = 0; i < s.length(); i++) {
            if (vowels[sChar[i]]) {
                vowelLen++;
            }
            if (i >= k && vowels[sChar[i - k]]) {
                vowelLen--;
            }
            if (vowelLen == k) {
                return k;
            }
            ans = Math.max(ans, vowelLen);
        }
        return ans;
    }

[chocolate-emperor]

class Solution {
public:
    bool isVowel(char x){
        return (x=='a'|| x == 'e' || x == 'i' || x =='o' || x=='u');
    }
    int maxVowels(string s, int k) {
        int len = s.length();
        int l = 0, cnt = 0,MaxVowel = 0;

        for(int r = 0; r<len;r++){
            if(isVowel(s[r]))  cnt+=1;
            //if (r-l+1<k)     continue;

            if (r-l+1>k){
                if(isVowel(s[l]))  cnt-=1;
                l+=1;
            }

            MaxVowel = max(MaxVowel,cnt);
        }
        return MaxVowel;
    }
};

[X1AOX1A]

1456. 定长子串中元音的最大数目

题目

给你字符串 s 和整数 k 。

请返回字符串 s 中长度为 k 的单个子字符串中可能包含的最大元音字母数。

英文中的 元音字母 为（a, e, i, o, u）。

示例 1：
输入：s = "abciiidef", k = 3
输出：3
解释：子字符串 "iii" 包含 3 个元音字母。

示例 2：
输入：s = "aeiou", k = 2
输出：2
解释：任意长度为 2 的子字符串都包含 2 个元音字母。

示例 3：
输入：s = "leetcode", k = 3
输出：2
解释："lee"、"eet" 和 "ode" 都包含 2 个元音字母。

示例 4：
输入：s = "rhythms", k = 4
输出：0
解释：字符串 s 中不含任何元音字母。

示例 5：
输入：s = "tryhard", k = 4
输出：1

思路：滑动窗口

• 维护一个 Out 和 In 指针，对窗口内的值进行更新

  

代码

class Solution:
    def maxVowels(self, s: str, k: int) -> int:
        if k>len(s): k = len(s)
        vowels = {"a", "e", "i", "o", "u"}
        # init
        cur = 0
        for i in range(k):
            if s[i] in vowels: cur += 1
        ans = cur
        for Out in range(0, len(s)-k):
            In = Out+k
            if s[Out] in vowels: cur -= 1
            if s[In] in vowels: cur += 1
            ans = max(ans, cur)
            # 达到上限，提前终止
            if ans==k: return k
        return ans

• 时间复杂度：O(n)
• 空间复杂度：O(1)

[wwewwt]

思路

双指针在字符串数组上移动，新增元音字母加1，出局元音字母减一

代码

class Solution:
    def maxVowels(self, s: str, k: int) -> int:
        i=0
        j=0
        aeiou_num = 0
        for j in range(k):
            if s[j] in ['a','e','i','o','u']:
                aeiou_num += 1

        i = 1
        j +=1
        max = aeiou_num
        while j < len(s):

            if s[i-1] in ['a','e','i','o','u']:
                aeiou_num -= 1

            if s[j] in ['a','e','i','o','u']:
                aeiou_num += 1

            if aeiou_num > max:
                max = aeiou_num

            if max == k:
                return k

            i+=1
            j+=1

        return max

复杂度

时间复杂度 O(n)，空间复杂度O(1)

[snmyj]

class Solution {
public:
    int maxVowels(string s, int k) {
        int n=s.size();
        int max=0;
        for(int i=0;i<n-k+1;i++){
            int cnt=0;
            for(int j=i;j<i+k;j++){
                if(s[j]=='a'||s[j]=='e'||s[j]=='i'||s[j]=='o'||s[j]=='u') cnt++;
                if(cnt==k) return k;
            }
            if(cnt>max) max=cnt;

        }
        return max;
    }
};

[yingchehu]

思路

固定長度滑動窗口

Code

class Solution:
    def maxVowels(self, s: str, k: int) -> int:
        r = l = 0
        ans = cnt = 0

        while r < len(s):
            if s[r] in ['a', 'e', 'i', 'o', 'u']:
                cnt += 1

            while l <= r:
                if r - l + 1 > k:
                    if s[l] in ['a', 'e', 'i', 'o', 'u']:
                        cnt -= 1
                    l += 1
                else:
                    break

            ans = max(ans, cnt)
            r += 1

        return ans

複雜度

Time: O(n) Space: O(1)

[Meisgithub]

class Solution {
public:
    int maxVowels(string s, int k) {
        int cnt = 0;
        int ans = 0;
        int n = s.size();
        for (int i = 0; i < k; ++i)
        {
            if (s[i] == 'a' || s[i] == 'e' || s[i] == 'i' || s[i] == 'o' || s[i] == 'u')
            {
                cnt++;
            }
        }
        ans = cnt;
        for (int i = k; i < n; ++i)
        {
            if (s[i - k] == 'a' || s[i - k] == 'e' || s[i - k] == 'i' || s[i - k] == 'o' || s[i - k] == 'u')
            {
                cnt--;
            }
            if (s[i] == 'a' || s[i] == 'e' || s[i] == 'i' || s[i] == 'o' || s[i] == 'u')
            {
                cnt++;
            }
            ans = max(ans, cnt);
        }
        return ans;
    }
};

[lp1506947671]

class Solution:
    def maxVowels(self, s: str, k: int) -> int:
        res = 0
        temp = 0
        vowels = set(['a','e','i','o','u'])
        for i in range(k):
            res += s[i] in vowels
        if res==k: return k
        temp = res
        for i in range(k,len(s)):
            temp += (s[i] in vowels) - (s[i-k] in vowels)
            res = max(temp,res)
            if res ==k: return k
        return res

复杂度分析

• 时间复杂度：O(n)，n 为字串长度
• 空间复杂度：O(1)

[kangliqi1]

class Solution { public int maxVowels(String s, int k) { int n = s.length(); int vowel_count = 0; for (int i = 0; i < k; ++i) { vowel_count += isVowel(s.charAt(i)); } int ans = vowel_count; for (int i = k; i < n; ++i) { vowel_count += isVowel(s.charAt(i)) - isVowel(s.charAt(i - k)); ans = Math.max(ans, vowel_count); } return ans; }

public int isVowel(char ch) {
    return ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u' ? 1 : 0;
}

}

[aoxiangw]

class Solution:
    def maxVowels(self, s: str, k: int) -> int:
        maxv, ans = 0, 0
        vowels = {'a', 'e', 'i', 'o', 'u'}
        for i, c in enumerate(s):
            if c in vowels:
                maxv += 1
            if i >= k and s[i - k] in vowels:
                maxv -= 1
            ans = max(ans, maxv)
        return ans

time, space: O(n), O(1)

[AstrKing]

public int maxVowels(String s, int k) {

if (s == null || s.length() < k)
    return 0;

int res = 0;
Set<Character> set = new HashSet<>(){{
    add('a');add('e');add('i');add('o');add('u');
}};

// init
for (int i = 0; i < k; i++)
    if (set.contains(s.charAt(i)))
        res++;

int cur = res;
for (int i = 1; i < s.length() - k + 1; i++) {

    if (set.contains(s.charAt(i - 1)))
        cur--;
    if (set.contains(s.charAt(i + k - 1)))
        cur++;

    res = Math.max(res, cur);
}

return res;

}

[CruiseYuGH]

"" class Solution: def maxVowels(self, s: str, k: int) -> int: current = 0 result = 0 for pos in range(k): if s[pos] in "aeiou": current += 1 result = current for pos in range(1, len(s) - k + 1): if s[pos - 1] in "aeiou": current -= 1 if s[pos + k - 1] in "aeiou": current += 1 result = max(result, current) return result ""

[RestlessBreeze]

code

class Solution {
public:
    int maxVowels(string s, int k) {
        int res = 0;
        for (int i = 0; i < k; i++)
        {
            if (s[i] == 'a' || s[i] == 'e' || s[i] == 'i' || s[i] == 'o' || s[i] == 'u')
                res++;
        }
        int endIndex = k;
        int startIndex = 1;
        int curres = res;
        for (; endIndex < s.length(); endIndex++, startIndex++)
        {
            if (s[startIndex - 1] == 'a' || s[startIndex - 1] == 'e' || s[startIndex - 1] == 'i' || s[startIndex - 1] == 'o' || s[startIndex - 1] == 'u')
                curres--;
            if (s[endIndex] == 'a' || s[endIndex] == 'e' || s[endIndex] == 'i' || s[endIndex] == 'o' || s[endIndex] == 'u')
                curres++;
            if (curres > res)
                res = curres;
        }
        return res;
    }
};

[KrabbeJing]

class Solution:
    def maxVowels(self, s: str, k: int) -> int:
        vowels = ['a', 'e', 'i', 'o', 'u']
        res = 0
        for i in range(k):
            res += 1 if s[i] in vowels else 0
        if res == k: return k
        temp = res
        for i in range(k, len(s)):
            temp += (s[i] in vowels) - (s[i-k] in vowels)
            if temp == k: return k
            res = max(temp, res)
        return res

[jiujingxukong]

思路

可以有以下几个思路

• 思路一暴力：题目要求我们找出所有 k 长度子串中可能包含的最大元音字母数，那我们遍历一边所有长度为 k 的子串不就知道啦。超时。
• 思路二：利用前缀和，只不过我们前缀和数组元素 i 存的是子串 0..i 的元音字母个数，这样再遍历一遍前缀和数组就可以求出结果 → 前缀和方案
• 思路三：维护一个窗口大小为 k 的滑窗即可，每移动一次可以归纳为：

1. 窗口左端弹出一个字符（删除步）
2. 若删除了元音则计数器-1（更新步）
3. 窗口右端加进来一个字符（添加步）
4. 若添加的字符是元音则计数器+1（更新步）

思路三代码

/**
 * @param {string} s
 * @param {number} k
 * @return {number}
 */
var maxVowels = function (s, k) {
  const dict = new Set(["a", "e", "i", "o", "u"]);
  let count = 0;
  for (let i = 0; i < k; i++) {
    if (dict.has(s[i])) count++;
  } 
  let temp = count;
  for (let i = k, j = 0; j < s.length; i++, j++) {
    if (dict.has(s[i])) temp++;

    if (dict.has(s[j])) temp--;

    count = Math.max(count, temp);
  }
  return count;
};

复杂度分析

TC:O(N).N为字符串长度 SC:O(1)

[Lydia61]

1456. 定长字串中元音的最大数目

思路

滑动窗口

代码

class Solution {
public:
    // 判断是否为元音
    bool isVowel(char ch) {
        return ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u'; 
    }

    int maxVowels(string s, int k) {
        int n = s.size();
        int vowel_count = 0;

        // 计算前第一个窗口的元音数量
        for (int i = 0; i < k; ++i) {
            vowel_count += isVowel(s[i]);
        }

        int ans = vowel_count;

        // 滑动窗口
        for (int i = k; i < n; ++i) {
            vowel_count += isVowel(s[i]) - isVowel(s[i - k]);
            ans = max(ans, vowel_count);
        }
        return ans;
    }
};

复杂度分析

• 时间复杂度：O(N)
• 空间复杂度：O(1)

[tzuikuo]

思路

待定

代码

class Solution {
public:
    bool isvowel(char c){
        return c=='a' or c=='e' or c=='i' or c=='o' or c=='u';
    }
    int maxVowels(string s, int k) {
        int i=0;
        int j=0;
        int n=s.size();
        int len=0;
        int ctv=0;
        int res=0;
        //Code starts
        while(j<n){
            if(isvowel(s[j])){
                ctv++;
            }
            len++;
            if(len>k){
                if(isvowel(s[i])){
                    ctv--;
                }
                i++;
            }
            res=max(res,ctv);
            j++;
        }
        return res;
    }
};

复杂度分析 待定

[joemonkeylee]

思路

滑动窗口

代码

           public int MaxVowels(string s, int k)
        {
            int count = 0;
            for (int i = 0; i < k; ++i)
            {
                count += isYuanYin(s[i]);
            }
            int ans = count;
            for (int i = k; i < s.Length; ++i)
            {
                count += isYuanYin(s[i]) - isYuanYin(s[i - k]);
                ans = Math.Max(ans, count);
            }
            return ans;
        }

        public int isYuanYin(char ch)
        {
            return ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u' ? 1 : 0;
        }

复杂度分析

• 时间复杂度：O(n)
• 空间复杂度：O(1)