根据其中的元音数量对字符串数组进行排序

给定N个字符串的数组arr [] ，任务是根据其中的元音数量对这些字符串进行排序。

例子：

Input: arr[] = { “geeks”, “for”, “coding” } 
Output: for, coding, geeks 
for -> o = 1 vowel 
coding -> o, i = 2 vowels 
geeks -> e, e = 2 vowels

Input: arr[] = { “lmno”, “pqrst”, “aeiou”, “xyz” } 
Output: pqrst, xyz, lmno, aeiou  

方法：想法是将每个元素及其元音数量存储在向量对中，然后根据存储的元音数量对向量中的所有元素进行排序。最后，按顺序打印字符串。

下面是上述方法的实现：

C++

// C++ implementation of the approach
 
#include 
using namespace std;
 
// Function to check the Vowel
bool isVowel(char ch)
{
    ch = toupper(ch);
    return (ch == 'A' || ch == 'E'
            || ch == 'I' || ch == 'O'
            || ch == 'U');
}
 
// Returns count of vowels in str
int countVowels(string str)
{
    int count = 0;
    for (int i = 0; i < str.length(); i++)
        if (isVowel(str[i])) // Check for vowel
            ++count;
    return count;
}
 
// Function to sort the array according to
// the number of the vowels
void sortArr(string arr[], int n)
{
    // Vector to store the number of vowels
    // with respective elements
    vector > vp;
 
    // Inserting number of vowels
    // with respective strings
    // in the vector pair
    for (int i = 0; i < n; i++) {
 
        vp.push_back(
            make_pair(
                countVowels(
                    arr[i]),
                arr[i]));
    }
 
    // Sort the vector, this will sort the pair
    // according to the number of vowels
    sort(vp.begin(), vp.end());
 
    // Print the sorted vector content
    for (int i = 0; i < vp.size(); i++)
        cout << vp[i].second << " ";
}
 
// Driver code
int main()
{
    string arr[] = { "lmno", "pqrst",
                     "aeiou", "xyz" };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    sortArr(arr, n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG{
 
static class pair
{
    int first;
    String second;
     
    pair(int first,String second)
    {
        this.first = first;
        this.second = second;
    }
}
 
// Function to check the Vowel
static boolean isVowel(char ch)
{
    ch = Character.toUpperCase(ch);
    return (ch == 'A' || ch == 'E' ||
            ch == 'I' || ch == 'O' ||
            ch == 'U');
}
 
// Returns count of vowels in str
static int countVowels(String str)
{
    int count = 0;
    for(int i = 0; i < str.length(); i++)
     
        // Check for vowel
        if (isVowel(str.charAt(i)))
            ++count;
             
    return count;
}
 
// Function to sort the array according to
// the number of the vowels
static void sortArr(String arr[], int n)
{
     
    // Vector to store the number of vowels
    // with respective elements
    ArrayList vp = new ArrayList<>();
 
    // Inserting number of vowels
    // with respective strings
    // in the vector pair
    for(int i = 0; i < n; i++)
    {
        vp.add(new pair(countVowels(arr[i]),
                                    arr[i]));
    }
 
    // Sort the vector, this will sort the pair
    // according to the number of vowels
    Collections.sort(vp, (a, b) -> a.first - b.first);
     
    // Print the sorted vector content
    for(int i = 0; i < vp.size(); i++)
        System.out.print(vp.get(i).second + " ");
}
 
// Driver code
public static void main(String[] args)
{
    String arr[] = { "lmno", "pqrst",
                     "aeiou", "xyz" };
    int n = arr.length;
     
    sortArr(arr, n);
}
}
 
// This code is contributed by offbeat

Python3

# Python3 implementation of the approach
 
# Function to check the Vowel
def isVowel(ch) :
     
    ch = ch.upper();
    return (ch == 'A' or ch == 'E'or ch == 'I' or
            ch == 'O'or ch == 'U');
 
# Returns count of vowels in str
def countVowels(string) :
 
    count = 0;
    for i in range(len(string)) :
         
        # Check for vowel
        if (isVowel(string[i])) :
            count += 1;
             
    return count;
 
# Function to sort the array according to
# the number of the vowels
def sortArr(arr, n) :
     
    # Vector to store the number of vowels
    # with respective elements
    vp = [];
 
    # Inserting number of vowels
    # with respective strings
    # in the vector pair
    for i in range(n) :
 
        vp.append((countVowels(arr[i]),arr[i]));
 
    # Sort the vector, this will sort the pair
    # according to the number of vowels
    vp.sort()
 
    # Print the sorted vector content
    for i in range(len(vp)) :
        print(vp[i][1], end= " ");
 
# Driver code
if __name__ == "__main__" :
    arr = [ "lmno", "pqrst","aeiou", "xyz" ];
    n = len(arr);
 
    sortArr(arr, n);
 
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
     
    // Function to check the Vowel
    static bool isVowel(char ch)
    {
        ch = char.ToUpper(ch);
        return (ch == 'A' || ch == 'E'
                || ch == 'I' || ch == 'O'
                || ch == 'U');
    }
      
    // Returns count of vowels in str
    static int countVowels(string str)
    {
        int count = 0;
        for (int i = 0; i < str.Length; i++)
            if (isVowel(str[i])) // Check for vowel
                ++count;
        return count;
    }
      
    // Function to sort the array according to
    // the number of the vowels
    static void sortArr(string[] arr, int n)
    {
       
        // Vector to store the number of vowels
        // with respective elements
        List> vp = new List>();
      
        // Inserting number of vowels
        // with respective strings
        // in the vector pair
        for (int i = 0; i < n; i++)
        { 
            vp.Add(new Tuple(countVowels(arr[i]), arr[i]));
        }
      
        // Sort the vector, this will sort the pair
        // according to the number of vowels
        vp.Sort();
      
        // Print the sorted vector content
        for (int i = 0; i < vp.Count; i++)
            Console.Write(vp[i].Item2 + " ");
    }
      
  // Driver code
  static void Main()
  {
    string[] arr = { "lmno", "pqrst",
                     "aeiou", "xyz" };
    int n = arr.Length;
    sortArr(arr, n);
  }
}
 
// This code is contributed by divyesh072019

输出：

pqrst xyz lmno aeiou

时间复杂度： O(N * log N)