对字母数字字符串进行排序，以使字母和数字的位置保持不变

本文概述

• C++
• Java
• Python3
• C#

给定一个字母数字字符串str, 任务是按以下方式对字符串进行排序：如果某个位置被字母占据, 那么它必须在排序后被一个字母占据；如果一个位置被数字占据, 则在排序后必须被一个数字占据。

例子：

输入：str =" geeks12for32geeks"
输出：eeeef12ggk23korss
输入：str =" d4c3b2a1"
输出：a1b2c3d4

方法: 将字符串转换为字符数组，然后对字符数组c[]进行排序。对字符数组进行排序后，数字字符将占据数组的起始索引，字母将占据数组的其余部分。

数字部分将被排序，字母部分也将被排序。我们将保持两个指标在al_c字母表的起始索引部分和一个nu_c开始指数的数字部分,现在我们将检查原始字符串,如果被一个字母然后占领我们将取代它与c [al_c]和增量al_c其他我们将取代它与c [nu_c]和增量nu_c。

下面是上述方法的实现：

C++

//C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
//Function that returns the string s
//in sorted form such that the
//positions of alphabets and numeric
//digits remain unchanged
string sort(string s)
{
     char c[s.length() + 1];
  
     //String to character array
     strcpy (c, s.c_str());
  
     //Sort the array
     sort(c, c + s.length());
  
     //Count of alphabets and numbers
     int al_c = 0, nu_c = 0;
  
     //Get the index from where the
     //alphabets start
     while (c[al_c] <97)
         al_c++;
  
     //Now replace the string with sorted string
     for ( int i = 0; i <s.length(); i++) {
  
         //If the position was occupied by an
         //alphabet then replace it with alphabet
         if (s[i] <97)
             s[i] = c[nu_c++];
  
         //Else replace it with a number
         else
             s[i] = c[al_c++];
     }
  
     //Return the sorted string
     return s;
}
  
//Driver code
int main()
{
     string s = "d4c3b2a1" ;
  
     cout <<sort(s);
  
     return 0;
}

Java

//A Java implementation of the approach
import java.util.*;
  
class GFG 
{
  
//Function that returns the string s
//in sorted form such that the
//positions of alphabets and numeric
//digits remain unchanged
static String sort(String s)
{
     char []c = new char [s.length() + 1 ];
  
     //String to character array
     c = s.toCharArray();
  
     //Sort the array
     Arrays.sort(c);
  
     //Count of alphabets and numbers
     int al_c = 0 , nu_c = 0 ;
  
     //Get the index from where the
     //alphabets start
     while (c[al_c] <97 )
         al_c++;
  
     //Now replace the string with sorted string
     for ( int i = 0 ; i <s.length(); i++)
     {
  
         //If the position was occupied by an
         //alphabet then replace it with alphabet
         if (s.charAt(i) <97 )
             s = s.substring( 0 , i)+ c[nu_c++]+s.substring(i+ 1 );
  
         //Else replace it with a number
         else
             s = s.substring( 0 , i)+ c[al_c++]+s.substring(i+ 1 );
     }
  
     //Return the sorted string
     return s;
}
  
//Driver code
public static void main(String[] args)
{
     String s = "d4c3b2a1" ;
  
     System.out.println(sort(s));
}
}
  
/* This code contributed by PrinciRaj1992 */

Python3

# Python3 implementation of the approach
  
# Function that returns the string s
# in sorted form such that the
# positions of alphabets and numeric
# digits remain unchanged
def sort(s):
  
     # String to character array
     c, s = list (s), list (s)
  
     # Sort the array
     c.sort()
  
     # Count of alphabets and numbers
     al_c = 0
     nu_c = 0
  
     # Get the index from where the
     # alphabets start
     while ord (c[al_c]) <97 :
         al_c + = 1
  
     # Now replace the string with sorted string
     for i in range ( len (s)):
  
         # If the position was occupied by an
         # alphabet then replace it with alphabet
         if s[i] <'a' :
             s[i] = c[nu_c]
             nu_c + = 1
  
         # Else replace it with a number
         else :
             s[i] = c[al_c]
             al_c + = 1
  
     # Return the sorted string
     return ''.join(s)
  
# Driver Code
if __name__ = = "__main__" :
     s = "d4c3b2a1"
     print (sort(s))
  
# This code is contributed by
# sanjeev2552

C#

//C# implementation of the approach 
using System;
  
class GFG 
{ 
  
     //Function that returns the string s 
     //in sorted form such that the 
     //positions of alphabets and numeric 
     //digits remain unchanged 
     static string sort( string s) 
     { 
         char []c = new char [s.Length + 1]; 
      
         //String to character array 
         c = s.ToCharArray(); 
      
         //Sort the array 
         Array.Sort(c); 
      
         //Count of alphabets and numbers 
         int al_c = 0, nu_c = 0; 
      
         //Get the index from where the 
         //alphabets start 
         while (c[al_c] <97) 
             al_c++; 
      
         //Now replace the string with sorted string 
         for ( int i = 0; i <s.Length; i++) 
         { 
      
             //If the position was occupied by an 
             //alphabet then replace it with alphabet 
             if (s[i] <97) 
                 s = s.Substring(0, i)+ c[nu_c++]+s.Substring(i+1); 
      
             //Else replace it with a number 
             else
                 s = s.Substring(0, i)+ c[al_c++]+s.Substring(i+1); 
         } 
      
         //Return the sorted string 
         return s; 
     } 
      
     //Driver code 
     public static void Main() 
     { 
         string s = "d4c3b2a1" ; 
      
         Console.WriteLine(sort(s)); 
     } 
} 
  
/* This code contributed by AnkitRai01 */

输出如下：

a1b2c3d4

时间复杂度：O(N * log(N))其中N是字符串的长度。

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