字符的字符串排序

给定字符串从“ a”到“ z”的小写字符。我们需要编写一个程序以按顺序打印此字符串的字符。

例子：

Input : bbccdefbbaa 
Output : aabbbbccdef

Input : geeksforgeeks
Output : eeeefggkkorss

一种简单的方法是使用排序算法，例如快速排序或合并排序，并对输入字符串排序并打印出来。

C++

// C++ program to sort a string of characters
#include
using namespace std;
  
// function to print string in sorted order
void sortString(string &str)
{
   sort(str.begin(), str.end());
   cout << str;
}
  
// Driver program to test above function
int main()
{
    string s = "geeksforgeeks"; 
    sortString(s); 
    return 0;
}

Java

// Java program to sort a string of characters 
  
import java.util.Arrays;
  
class GFG {
  
// function to print string in sorted order 
    static void sortString(String str) {
        char []arr = str.toCharArray();
        Arrays.sort(arr);
        System.out.print(String.valueOf(arr));
    }
  
// Driver program to test above function 
    public static void main(String[] args) {
        String s = "geeksforgeeks";
        sortString(s);
    }
}
// This code is contributed by Rajput-Ji

Python3

# Python3 program to sort a string 
# of characters 
  
# function to print string in
# sorted order 
def sortString(str) :
    str = ''.join(sorted(str))
    print(str)
  
# Driver Code
s = "geeksforgeeks"
sortString(s) 
  
# This code is contributed by Smitha

C#

// C# program to sort a string of characters 
using System;    
public class GFG { 
  
// function to print string in sorted order 
    static void sortString(String str) { 
        char []arr = str.ToCharArray(); 
        Array.Sort(arr); 
        Console.WriteLine(String.Join("",arr)); 
    } 
  
// Driver program to test above function 
    public static void Main() { 
        String s = "geeksforgeeks"; 
        sortString(s); 
    } 
} 
// This code is contributed by 29AjayKumar

---

Python3

# Python 3 program to sort a string 
# of characters
  
MAX_CHAR = 26
  
# function to print string in sorted order
def sortString(str):
      
    # Hash array to keep count of characters.
    # Initially count of all charters is 
    # initialized to zero.
    charCount = [0 for i in range(MAX_CHAR)]
      
    # Traverse string and increment 
    # count of characters
    for i in range(0, len(str), 1):
          
        # 'a'-'a' will be 0, 'b'-'a' will be 1,
        # so for location of character in count 
        # array we wil do str[i]-'a'.
        charCount[ord(str[i]) - ord('a')] += 1
      
    # Traverse the hash array and print 
    # characters
    for i in range(0, MAX_CHAR, 1):
        for j in range(0, charCount[i], 1):
            print(chr(ord('a') + i), end = "")
  
# Driver Code
if __name__ == '__main__':
    s = "geeksforgeeks"
    sortString(s)
      
# This code is contributed by
# Sahil_Shelangia
.

输出：

eeeefggkkorss

时间复杂度： O(n log n)，其中n是字符串的长度。

一种有效的方法是首先观察到总共只能有26个唯一字符。因此，我们可以将所有字符从’a’到’z’的出现次数存储在哈希数组中。哈希数组的第一个索引将代表字符“ a”，第二个索引将代表“ b”，依此类推。最后，我们将简单地遍历哈希数组，并将字符从’a’打印到’z’，将其出现在输入字符串。

下面是上述想法的实现：

C++

// C++ program to sort a string of characters
#include
using namespace std;
  
const int MAX_CHAR = 26;
  
// function to print string in sorted order
void sortString(string &str)
{
    // Hash array to keep count of characters.
    // Initially count of all charters is 
    // initialized to zero.
    int charCount[MAX_CHAR] = {0};
      
    // Traverse string and increment 
    // count of characters
    for (int i=0; i

Java

// Java program to sort
// a string of characters
public class SortString{
    static final int MAX_CHAR = 26;
  
    // function to print string in sorted order
    static void sortString(String str) {
  
        // Hash array to keep count of characters.
        int letters[] = new int[MAX_CHAR];
  
        // Traverse string and increment
        // count of characters
        for (char x : str.toCharArray()) {
  
            // 'a'-'a' will be 0, 'b'-'a' will be 1,
            // so for location of character in count
            // array we will do str[i]-'a'.
            letters[x - 'a']++;
        }
  
        // Traverse the hash array and print
        // characters
        for (int i = 0; i < MAX_CHAR; i++) {
            for (int j = 0; j < letters[i]; j++) {
                System.out.print((char) (i + 'a'));
            }
        }
    }
  
    // Driver program to test above function
    public static void main(String[] args) {
        sortString("geeksforgeeks");
    }
}
// This code is contributed 
// by Sinuhe

Python3

# Python 3 program to sort a string 
# of characters
  
MAX_CHAR = 26
  
# function to print string in sorted order
def sortString(str):
      
    # Hash array to keep count of characters.
    # Initially count of all charters is 
    # initialized to zero.
    charCount = [0 for i in range(MAX_CHAR)]
      
    # Traverse string and increment 
    # count of characters
    for i in range(0, len(str), 1):
          
        # 'a'-'a' will be 0, 'b'-'a' will be 1,
        # so for location of character in count 
        # array we wil do str[i]-'a'.
        charCount[ord(str[i]) - ord('a')] += 1
      
    # Traverse the hash array and print 
    # characters
    for i in range(0, MAX_CHAR, 1):
        for j in range(0, charCount[i], 1):
            print(chr(ord('a') + i), end = "")
  
# Driver Code
if __name__ == '__main__':
    s = "geeksforgeeks"
    sortString(s)
      
# This code is contributed by
# Sahil_Shelangia

。

C＃

// C# program to sort 
// a string of characters 
using System;
  
class GFG
{
      
    // Method to sort a 
    // string alphabetically 
    public static string sortString(string inputString)
    {
          
        // convert input 
        // string to char array 
        char[] tempArray = inputString.ToCharArray();
  
        // sort tempArray 
        Array.Sort(tempArray);
  
        // return new sorted string 
        return new string(tempArray);
    }
  
    // Driver Code 
    public static void Main(string[] args)
    {
        string inputString = "geeksforgeeks";
  
        Console.WriteLine(sortString(inputString));
    }
}
  
// This code is contributed by Shrikant13

输出：

eeeefggkkorss

时间复杂度： O(n)，其中n是输入字符串的长度。
辅助空间： O(1)。