根据字符的 ASCII 值对字符串进行排序

给定一个大小为N的字符串S ，任务是根据字符串的 ASCII 值对字符串进行排序。

例子：

Input: S = “Geeks7”
Output: 7Geeks
Explanation: According to the ASCII values, integers comes first, then capital alphabets and the small alphabets.

Input: S = “GeeksForGeeks”
Output: FGGeeeekkorss

方法：解决这个问题的想法是维护一个数组来存储每个字符的频率，然后将它们相应地添加到结果字符串中。由于最多有256个字符，这使得空间复杂度保持不变。请按照以下步骤解决此问题：

• 用值0初始化大小为256的向量freq[]以存储字符串中每个字符的频率。
• 使用变量i遍历范围[0, N)并将数组freq[]中s[i]的计数增加1 。
• 将字符串S设为空字符串。
• 使用变量 i 迭代范围[0, N)并使用变量j迭代范围 [0, freq[ i ])并添加对应于字符s[]中第 i 个ASCII 值的字符串。
• 执行上述步骤后，打印字符串S作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to sort the string based
// on their ASCII values
void sortString(string s)
{
    int N = s.length();
 
    // Stores the frequency of each
    // character of the string
    vector freq(256, 0);
 
    // Count and store the frequency
    for (int i = 0; i < N; i++) {
        freq[s[i]]++;
    }
 
    s = "";
 
    // Store the result
    for (int i = 0; i < 256; i++) {
        for (int j = 0; j < freq[i]; j++)
            s = s + (char)i;
    }
 
    // Print the result
    cout << s << "\n";
 
    return;
}
 
// Driver Code
int main()
{
    string S = "GeeksForGeeks";
    sortString(S);
 
    return 0;
}

Java

// java program for the above approach
import java.util.*;
 
class Main
{
 
// Function to sort the string based
// on their ASCII values
static void sortString(String s)
{
    int N = s.length();
 
    // Stores the frequency of each
    // character of the string
    int [] freq = new int [256];
    for(int i = 0; i < 256; i++){
        freq[i] = 0;
    }
 
    // Count and store the frequency
    for (int i = 0; i < N; i++) {
         char character = s.charAt(i);
         int val = (int) character;
        freq[val]++;
    }
 
    // Store the result
    for (int i = 0; i < 256; i++) {
        for (int j = 0; j < freq[i]; j++)
          //    s = s + (char)i;
            System.out.print((char)i);
    }
}
 
// Driver Code
public static void main(String [] args)
{
    String S = "GeeksForGeeks";
    sortString(S);
}
}
 
// This code is contributed by amreshkumar3.

Python3

# Python Program to implement
# the above approach
 
# Function to sort the string based
# on their ASCII values
def sortString(s):
    N = len(s)
 
    # Stores the frequency of each
    # character of the string
    freq = [0] * 256
 
    # Count and store the frequency
    for i in range(0, N) :
        freq[ord(s[i])] += 1
 
    s = ""
 
    # Store the result
    for i in range(256):
        for j in range(freq[i]):
            s = s + chr(i)
 
    # Print the result
    print(s)
 
    return
 
# Driver Code
S = "GeeksForGeeks"
sortString(S)
 
# This code is contributed by gfgking.

C#

// C# implementation for the above approach
using System;
class GFG
{
 
// Function to sort the string based
// on their ASCII values
static void sortString(string s)
{
    int N = s.Length;
 
    // Stores the frequency of each
    // character of the string
    int [] freq = new int[256];
    for(int i = 0; i < 256; i++){
        freq[i] = 0;
    }
 
    // Count and store the frequency
    for (int i = 0; i < N; i++) {
         char character = s[i];
         int val = (int) character;
        freq[val]++;
    }
 
    // Store the result
    for (int i = 0; i < 256; i++) {
        for (int j = 0; j < freq[i]; j++)
          //    s = s + (char)i;
           Console.Write((char)i);
    }
}
 
    // Driver Code
    public static void Main()
    {
        string S = "GeeksForGeeks";
        sortString(S);
    }
}
 
// This code is contributed by sanjoy_62.

Javascript

       // JavaScript Program to implement
       // the above approach
 
       // Function to sort the string based
       // on their ASCII values
       function sortString(s)
       {
           let N = s.length;
 
           // Stores the frequency of each
           // character of the string
           let freq = new Array(256).fill(0)
 
           // Count and store the frequency
           for (let i = 0; i < N; i++) {
               freq[s[i].charCodeAt(0)]++;
           }
 
           s = "";
 
           // Store the result
           for (let i = 0; i < 256; i++) {
               for (let j = 0; j < freq[i]; j++)
                   s = s + String.fromCharCode(i);
           }
 
           // Print the result
           document.write(s);
 
           return;
       }
 
       // Driver Code
       let S = "GeeksForGeeks";
       sortString(S);
 
       // This code is contributed by Potta Lokesh
 
   

输出：

FGGeeeekkorss

时间复杂度： O(256*N)
辅助空间： O(256)

替代方法：给定的问题也可以通过使用带有内置 sort()函数的比较器函数来解决，根据它们的 ASCII 值对给定的字符串进行排序。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include "bits/stdc++.h"
using namespace std;
 
// Comparator Function to sort the given
// string in increasing order of their
// ASCII value
bool cmp(char ch, char chh) { return int(ch) <= int(chh); }
 
// Function to sort the string based
// on their ASCII values
string sortString(string S)
{
    // Sort the string S
    sort(S.begin(), S.end(), cmp);
 
    return S;
}
 
// Driver Code
int main()
{
    string S = "GeeksForGeeks";
    cout << sortString(S);
 
    return 0;
}

Python3

# Python Program to implement
# the above approach
 
# Function to sort the string based
# on their ASCII values
def sortString(s):
    N = len(s)
 
    # Stores the frequency of each
    # character of the string
    freq = [0] * 256
 
    # Count and store the frequency
    for i in range(0, N) :
        freq[ord(s[i])] += 1
 
    s = ""
 
    # Store the result
    for i in range(256):
        for j in range(freq[i]):
            s = s + chr(i)
 
    # Print the result
    print(s)
 
    return
 
# Driver Code
S = "GeeksForGeeks"
sortString(S)
 
# This code is contributed by code_hunt.

输出：

FGGeeeekkorss

时间复杂度： O(N*log N)
辅助空间： O(1)