按出现顺序打印偶数频率的字符
原文：https://www.geeksforgeeks.org/print-characters-having-even-frequencies-in-order-of-occurrence/
给定仅包含小写字符的字符串str。 任务是按照出现的顺序打印偶数频率的字符。
注意：重复出现的偶数频率按出现的次数打印。
示例：
输入：
str = "geeksforgeeks"输出：
geeksgeeks
字符 频率 g2 e4 k2 s2 f1 o1 r1 仅有
g，e，k和s是频率偶数的字符。输入：
str = "aeroplane"输出：
aeae
方法：创建一个频率数组，以存储给定字符串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 <bits/stdc++.h>
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
输出：
geeksgeeks
时间复杂度：O(n)。
辅助空间：O(1)。
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