按出现顺序打印奇数频率的字符
原文:https://www.geeksforgeeks.org/print-characters-having-odd-frequencies-in-order-of-occurrence/
给定仅包含小写字符的字符串str
。 任务是按照出现的频率打印出具有奇数频率的字符。
注意:具有奇数频率的重复元素按出现的次数打印多次。
示例:
输入:
str = "geeksforgeeks"
输出:
for
字符 频率 g
2 e
4 k
2 s
2 f
1 o
1 r
1
f
,o
和r
是唯一具有奇数频率的字符。输入:
str = "elephant"
输出:
lphant
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奇数。 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 odd 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 odd
if (freq[str[i] - 'a'] % 2 == 1) {
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
class GFG {
// Function to print the odd frequency characters
// in the order of their occurrence
public static void printChar(String str, int n)
{
// To store the frequency of each of
// the character of the string
int[] freq = new int[26];
// 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 odd
if (freq[str.charAt(i) - 'a'] % 2 == 1) {
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 Naman_Garg.
Python3
# Python3 implementation of the approach
import sys
import math
# Function to print the odd frequency characters
# in the order of their occurrence
def printChar(str_, n):
# To store the frequency of each of
# the character of the string and
# Initialize all elements of freq[] to 0
freq = [0] * 26
# Update the frequency of each character
for i in range(n):
freq[ord(str_[i]) - ord('a')] += 1
# Traverse str character by character
for i in range(n):
# If frequency of current character is odd
if (freq[ord(str_[i]) -
ord('a')]) % 2 == 1:
print("{}".format(str_[i]), end = "")
# Driver code
if __name__=='__main__':
str_ = "geeksforgeeks"
n = len(str_)
printChar(str_, n)
# This code is contributed by Vikash Kumar 37
C
// C# implementation of the approach
using System;
class GFG {
// Function to print the odd frequency characters
// in the order of their occurrence
public static void printChar(String str, int n)
{
// To store the frequency of each of
// the character of the string
int[] freq = new int[26];
// 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 odd
if (freq[str[i] - 'a'] % 2 == 1) {
Console.Write(str[i]);
}
}
}
// Driver code
public static void Main(String[] args)
{
String str = "geeksforgeeks";
int n = str.Length;
printChar(str, n);
}
}
// This code has been contributed by 29AjayKumar
输出:
for
时间复杂度:O(n)
。
辅助空间:O(1)
。
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