按出现顺序打印具有偶数频率的字符

给定一个只包含小写字符的字符串str 。任务是按出现的顺序打印具有偶数频率的字符。
注意：具有偶数频率的重复元素按照它们出现的顺序打印它们出现的次数。
例子：

Input: str = “geeksforgeeks” 
Output: geeksgeeks 
 

Character	Frequency
‘g’	2
‘e’	4
‘k’	2
‘s’	2
‘f’	1
‘o’	1
‘r’	1

‘g’, ‘e’, ‘k’ and ‘s’ are the only characters with even frequencies.
Input: str = “aeroplane” 
Output: aeae 
 

方法：创建一个频率数组来存储给定字符串str的每个字符的频率。再次遍历字符串str并检查该字符的频率是否为偶数。如果是，则打印字符。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
#define SIZE 26
 
// Function to print the even frequency characters
// in the order of their occurrence
void printChar(string str, int n)
{
 
    // To store the frequency of each of
    // the character of the string
    int freq[SIZE];
 
    // Initialize all elements of freq[] to 0
    memset(freq, 0, sizeof(freq));
 
    // Update the frequency of each character
    for (int i = 0; i < n; i++)
        freq[str[i] - 'a']++;
 
    // Traverse str character by character
    for (int i = 0; i < n; i++) {
 
        // If frequency of current character is even
        if (freq[str[i] - 'a'] % 2 == 0) {
            cout << str[i];
        }
    }
}
 
// Driver code
int main()
{
    string str = "geeksforgeeks";
    int n = str.length();
    printChar(str, n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
static int SIZE = 26;
 
// Function to print the even frequency characters
// in the order of their occurrence
static void printChar(String str, int n)
{
 
    // To store the frequency of each of
    // the character of the string
    int []freq = new int[SIZE];
 
    // Update the frequency of each character
    for (int i = 0; i < n; i++)
        freq[str.charAt(i) - 'a']++;
 
    // Traverse str character by character
    for (int i = 0; i < n; i++)
    {
 
        // If frequency of current character is even
        if (freq[str.charAt(i) - 'a'] % 2 == 0)
        {
            System.out.print(str.charAt(i));
        }
    }
}
 
// Driver code
public static void main(String[] args)
{
    String str = "geeksforgeeks";
    int n = str.length();
    printChar(str, n);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
SIZE = 26
 
# Function to print the even frequency characters
# in the order of their occurrence
def printChar(string, n):
 
    # To store the frequency of each of
    # the character of the stringing
    # Initialize all elements of freq[] to 0
    freq = [0] * SIZE
 
    # Update the frequency of each character
    for i in range(0, n):
        freq[ord(string[i]) - ord('a')] += 1
 
    # Traverse string character by character
    for i in range(0, n):
 
        # If frequency of current character is even
        if (freq[ord(string[i]) -
                 ord('a')] % 2 == 0):
            print(string[i], end = "")
         
# Driver code
if __name__ == '__main__':
    string = "geeksforgeeks"
    n = len(string)
    printChar(string, n)
 
# This code is contributed by Ashutosh450

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
     
class GFG
{
static int SIZE = 26;
 
// Function to print the even frequency characters
// in the order of their occurrence
static void printChar(String str, int n)
{
 
    // To store the frequency of each of
    // the character of the string
    int []freq = new int[SIZE];
 
    // Update the frequency of each character
    for (int i = 0; i < n; i++)
        freq[str[i] - 'a']++;
 
    // Traverse str character by character
    for (int i = 0; i < n; i++)
    {
 
        // If frequency of current character is even
        if (freq[str[i] - 'a'] % 2 == 0)
        {
            Console.Write(str[i]);
        }
    }
}
 
// Driver code
public static void Main(String[] args)
{
    String str = "geeksforgeeks";
    int n = str.Length;
    printChar(str, n);
}
}
 
// This code is contributed by Princi Singh

Javascript

 
    // Javascript implementation of the approach
     
    let SIZE = 26;
     
    // Function to print the even frequency characters
    // in the order of their occurrence
    function printChar(str, n)
    {
        // To store the frequency of each of
        // the character of the string
        let freq = new Array(SIZE);
       
        // Initialize all elements of freq[] to 0
        for(let i = 0; i < freq.length; i++)
        {
            freq[i] = 0;
        }
       
        // Update the frequency of each character
        for (let i = 0; i < n; i++)
        {
            freq[str.charCodeAt(i) - 'a'.charCodeAt(0)]++;
        }
             
        // Traverse str character by character
        for (let i = 0; i < n; i++) {
       
            // If frequency of current character is even
            if (freq[str.charCodeAt(i) - 'a'.charCodeAt(0)] % 2 == 0) {
                document.write(str[i]);
            }
        }
    }
 
     
    let str = "geeksforgeeks";
    let n = str.length;
    printChar(str, n);
 

输出：

geeksgeeks

时间复杂度： O(n)
辅助空间： O(1)

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