最长的元音子串

给定一个由小写字母组成的字符串s，我们需要找到仅包含 (a, e, i, o, u) 的最长子串长度。

例子 ：

Input: s = "geeksforgeeks"
Output: 2
Longest substring is "ee"

Input: s = "theeare"
Output: 3

这个想法是遍历字符串并跟踪字符串中当前元音的数量。如果我们看到一个不是元音的字符，我们将 count 重置为 0。但在重置之前，我们会更新 max count 值，这将是我们的结果。

C++

// CPP program to find the
// longest substring of vowels.
#include 
using namespace std;
 
bool isVowel(char c)
{
    return (c == 'a' || c == 'e' || c == 'i'
            || c == 'o' || c == 'u');
            
}
 
int longestVowel(string s)
{
    int count = 0, res = 0;
    for (int i = 0; i < s.length(); i++)
    {
 
        // Increment current count
        // if s[i] is vowel
        if (isVowel(s[i]))
        count++;    
 
        else
        {
 
            // check previous value
            // is greater then or not
            res = max(res, count);
 
            count = 0;
        }
    }
    return max(res, count);
}
 
// Driver code
int main()
{
    string s = "theeare";
    cout << longestVowel(s) << endl;
    return 0;
}

Java

// Java program to find the
// longest substring of vowels.
import java.util.*;
 
class GFG
{
 
    static boolean isVowel(char c)
    {
        return (c == 'a' || c == 'e' || c == 'i'
                 || c == 'o' || c == 'u');
    }
 
    static int longestVowel(String str)
    {
        int count = 0, res = 0;
        char[] s = str.toCharArray();
         
        for (int i = 0; i < s.length; i++)
        {
 
            // Increment current count
            // if s[i] is vowel
            if (isVowel(s[i]))
            count++;    
 
            else
            {
                // check previous value
                // is greater then or not
                res = Math.max(res, count);
 
                count = 0;
            }
        }
         
    return Math.max(res, count);
    }
 
// Driver code
public static void main (String[] args)
{
    String s = "theeare";
    System.out.println(longestVowel(s));
}
}
// This code is contributed by Mr. Somesh Awasthi

Python3

# Python3 program to find the
# longest substring of vowels.
def isVowel(c):
    return (c == 'a' or c == 'e' or
            c == 'i' or c == 'o' or
            c == 'u')          
 
def longestVowel(s):
    count, res = 0, 0
     
    for i in range(len(s)):
 
        # Increment current count
        # if s[i] is vowel
        if (isVowel(s[i])):
            count += 1 
 
        else:
        
            # check previous value
            # is greater then or not
            res = max(res, count)
 
            count = 0
    return max(res, count)
 
# Driver code
if __name__ == "__main__":
    s = "theeare"
    print (longestVowel(s))
    
# This code is contributed by Chitranayal

C#

// C# program to find the
// longest substring of vowels.
using System;
 
class GFG
{
 
    static bool isVowel(char c)
    {
        return (c == 'a' || c == 'e' || c == 'i' 
                || c == 'o' || c == 'u');
    }
 
    static int longestVowel(String str)
    {
        int count = 0, res = 0;
        char []s = str.ToCharArray();
         
        for (int i = 0; i < s.Length; i++)
        {
 
            // Increment current count
            // if s[i] is vowel
            if (isVowel(s[i]))
            count++;    
 
            else
            {
                // check previous value
                // is greater then or not
                res = Math.Max(res, count);
 
                count = 0;
            }
        }
         
    return Math.Max(res, count);
    }
 
// Driver code
public static void Main ()
    {
        String s = "theeare";
        Console.Write(longestVowel(s));
    }
}
 
// This code is contributed by nitin mittal

Javascript

 
// Javascript program to find the
// longest substring of vowels.
function isVowel(c)
{
    return (c == 'a' || c == 'e' ||
            c == 'i' || c == 'o' ||
            c == 'u');
}
 
function longestVowel(str)
{
    let count = 0, res = 0;
    let s = str.split("");
      
    for(let i = 0; i < s.length; i++)
    {
         
        // Increment current count
        // if s[i] is vowel
        if (isVowel(s[i]))
            count++;   
        else
        {
             
            // Check previous value
            // is greater then or not
            res = Math.max(res, count);
 
            count = 0;
        }
    }
    return Math.max(res, count);
}
 
// Driver code
let s = "theeare";
document.write(longestVowel(s));
 
// This code is contributed by avanitrachhadiya2155
     

输出

3

时间复杂度：O(n)

辅助空间：O(1)