按字母顺序打印每个字符的频率
给定一个字符串 str ，任务是按字母顺序打印 str 的每个字符的频率。 例:
输入: str = "aabccccddd" 输出: a2b1c4d3 由于已经按字母顺序排列，所以返回每个字符的字符频率 。 输入:str = " geeksforgeeks " 输出: e4f1g2k2o1r1s2
进场:
- 创建一个映射来存储给定字符串中每个字符的频率。
- 遍历字符串并检查字符是否出现在地图中。
- 如果字符不存在，将它以 1 作为初始值插入到地图中，否则将其频率增加 1。
- 最后，按字母顺序打印每个字符的频率。
以下是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
const int MAX = 26;
// Function to print the frequency
// of each of the characters of
// s in alphabetical order
void compressString(string s, int n)
{
// To store the frequency
// of the characters
int freq[MAX] = { 0 };
// Update the frequency array
for (int i = 0; i < n; i++) {
freq[s[i] - 'a']++;
}
// Print the frequency in alphatecial order
for (int i = 0; i < MAX; i++) {
// If the current alphabet doesn't
// appear in the string
if (freq[i] == 0)
continue;
cout << (char)(i + 'a') << freq[i];
}
}
// Driver code
int main()
{
string s = "geeksforgeeks";
int n = s.length();
compressString(s, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG
{
static int MAX = 26;
// Function to print the frequency
// of each of the characters of
// s in alphabetical order
static void compressString(String s, int n)
{
// To store the frequency
// of the characters
int freq[] = new int[MAX] ;
// Update the frequency array
for (int i = 0; i < n; i++)
{
freq[s.charAt(i) - 'a']++;
}
// Print the frequency in alphatecial order
for (int i = 0; i < MAX; i++)
{
// If the current alphabet doesn't
// appear in the string
if (freq[i] == 0)
continue;
System.out.print((char)(i + 'a') +""+ freq[i]);
}
}
// Driver code
public static void main (String[] args)
{
String s = "geeksforgeeks";
int n = s.length();
compressString(s, n);
}
}
// This code is contributed by AnkitRai01
Python 3
# Python3 implementation of the approach
MAX = 26;
# Function to print the frequency
# of each of the characters of
# s in alphabetical order
def compressString(s, n) :
# To store the frequency
# of the characters
freq = [ 0 ] * MAX;
# Update the frequency array
for i in range(n) :
freq[ord(s[i]) - ord('a')] += 1;
# Print the frequency in alphatecial order
for i in range(MAX) :
# If the current alphabet doesn't
# appear in the string
if (freq[i] == 0) :
continue;
print((chr)(i + ord('a')),freq[i],end = " ");
# Driver code
if __name__ == "__main__" :
s = "geeksforgeeks";
n = len(s);
compressString(s, n);
# This code is contributed by AnkitRai01
C
// C# implementation of the approach
using System;
class GFG
{
static int MAX = 26;
// Function to print the frequency
// of each of the characters of
// s in alphabetical order
static void compressString(string s, int n)
{
// To store the frequency
// of the characters
int []freq = new int[MAX] ;
// Update the frequency array
for (int i = 0; i < n; i++)
{
freq[s[i] - 'a']++;
}
// Print the frequency in alphatecial order
for (int i = 0; i < MAX; i++)
{
// If the current alphabet doesn't
// appear in the string
if (freq[i] == 0)
continue;
Console.Write((char)(i + 'a') +""+ freq[i]);
}
}
// Driver code
public static void Main()
{
string s = "geeksforgeeks";
int n = s.Length;
compressString(s, n);
}
}
// This code is contributed by AnkitRai01
java 描述语言
<script>
// Javascript implementation of the approach
let MAX = 26;
// Function to print the frequency
// of each of the characters of
// s in alphabetical order
function compressString(s, n)
{
// To store the frequency
// of the characters
let freq = new Array(MAX);
freq.fill(0);
// Update the frequency array
for (let i = 0; i < n; i++)
{
freq[s[i].charCodeAt() - 'a'.charCodeAt()]++;
}
// Print the frequency in alphatecial order
for (let i = 0; i < MAX; i++)
{
// If the current alphabet doesn't
// appear in the string
if (freq[i] == 0)
continue;
document.write(String.fromCharCode(i +
'a'.charCodeAt()) +""+ freq[i]);
}
}
let s = "geeksforgeeks";
let n = s.length;
compressString(s, n);
</script>
Output:
e4f1g2k2o1r1s2
时间复杂度:O(n) T3】辅助空间: O(1)
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