按给定优先级的升序对字符串进行排序

给定长度为N的字母数字字符串S ，任务是根据以下条件以优先级从高到低的顺序对字符串进行排序：

• 甚至ASCII值的字符比奇数ASCII字符值更高的优先级。
• 偶数位比奇数位具有更高的优先级。
• 数字比字符具有更高的优先级。
• 对于具有相同奇偶校验的字符或数字，优先级按其ASCII值的升序排列。

例子：

Input: S = “abcd1234”
Output: 1324bdac
Explanation: 
The ASCII value of “a” is 97.
The ASCII value of “b” is 98.
The ASCII value of “c” is 99.
The ASCII value of “d” is 100.
Since characters with even ASCII value have higher priority, “b” and “d” are placed first followed by “a” and “c”.
Similarly, even digits have more priority than odd digits. Therefore, they are placed first.
Since the numbers have a higher priority than characters, the sorted string is “1324bdac”

Input: S = “adb123”
Output: bda132

方法：我们的想法是字符与奇数和偶数的ASCII值，也是数字与奇数和偶数校验分开。然后，按优先级顺序将这些子字符串连接起来。请按照以下步骤解决问题：

1. 初始化两个变量，例如数字和字符，以分别存储字符和数字。
2. 根据ASCII表对字符串数字和字符进行排序。
3. 现在，遍历所有字符中的字符，并且每个字符与一个奇数奇偶校验到可变发言权oddChars并用偶校验另一个变量比如evenChars每个字符追加。
4. 类似地，对于字符串数字，将奇数和偶数奇偶校验数字分开，例如oddDigs和evenDigs 。
5. 最后，将字符串连接为oddChars + evenChars + oddDigs + evenDigs 。

下面是上述方法的实现：

C++14

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to sort the string s
// based on the given conditions
string sortString(string s)
{
    // Stores digits of given string
    string digits = "";
 
    // Stores characters of given string
    string character = "";
 
    // Iterate over characters of the string
    for (int i = 0; i < s.size(); ++i) {
        if (s[i] >= '0' && s[i] <= '9')
 {
     digits += s[i];
 }
 else
 {
     character += s[i];
 }
 }
 
 // Sort the string of digits
 sort(digits.begin(), digits.end());
 
 // Sort the string of characters
 sort(character.begin(), character.end());
 
 // Stores odd and even characters
 string OddChar = "", EvenChar = "";
 
 // Seperate odd and even digits
 for (int i = 0; i < digits.length(); ++i) {
     if ((digits[i] - '0') % 2 == 1) {
         OddChar += digits[i];
     }
     else {
         EvenChar += digits[i];
     }
 }
 
 // Concatenate strings in the order
 // odd characters followed by even
 OddChar += EvenChar;
 
 EvenChar = "";
 
 // Seperate the odd and even chars
 for (int i = 0; i < character.length();
      ++i) {
 
     if ((character[i] - 'a') % 2 == 1) {
         OddChar += character[i];
     }
     else {
         EvenChar += character[i];
     }
 }
 
 // Final string
 OddChar += EvenChar;
 
 // Return the final string
 return OddChar;
 }
 
 // Driver Code
 int main()
 {
     // Given string
     string s = "abcd1234";
 
     // Returns the sorted string
     cout << sortString(s);
 
     return 0;
 }

Java

// Java program for the above approach
import java.util.*;
class GFG
{
 
    // Function to sort the String s
    // based on the given conditions
    static String sortString(String s)
    {
       
        // Stores digits of given String
        String digits = "";
 
        // Stores characters of given String
        String character = "";
 
        // Iterate over characters of the String
        for (int i = 0; i < s.length(); ++i) {
            if (s.charAt(i) >= '0' && s.charAt(i) <= '9') {
                digits += s.charAt(i);
            } else {
                character += s.charAt(i);
            }
        }
 
        // Sort the String of digits
        digits = sort(digits);
        // Sort the String of characters
        character = sort(character);
 
        // Stores odd and even characters
        String OddChar = "", EvenChar = "";
 
        // Seperate odd and even digits
        for (int i = 0; i < digits.length(); ++i) {
            if ((digits.charAt(i) - '0') % 2 == 1) {
                OddChar += digits.charAt(i);
            } else {
                EvenChar += digits.charAt(i);
            }
        }
 
        // Concatenate Strings in the order
        // odd characters followed by even
        OddChar += EvenChar;
        EvenChar = "";
 
        // Seperate the odd and even chars
        for (int i = 0; i < character.length(); ++i) {
 
            if ((character.charAt(i) - 'a') % 2 == 1) {
                OddChar += character.charAt(i);
            } else {
                EvenChar += character.charAt(i);
            }
        }
 
        // Final String
        OddChar += EvenChar;
 
        // Return the final String
        return OddChar;
    }
 
    static String sort(String inputString)
    {
       
        // convert input string to char array
        char tempArray[] = inputString.toCharArray();
 
        // sort tempArray
        Arrays.sort(tempArray);
 
        // return new sorted string
        return new String(tempArray);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given String
        String s = "abcd1234";
 
        // Returns the sorted String
        System.out.print(sortString(s));
 
    }
}
 
// This code is contributed by shikhasingrajput

Python3

# Python3 program for the above approach
 
# Function to sort the string s
# based on the given conditions
def sortString(s):
     
    # Stores digits of given string
    digits = ""
     
    # Stores characters of given string
    character = ""
     
    # Iterate over characters of the string
    for i in range(len(s)):
        if (s[i] >= '0' and s[i] <= '9'):
            digits += s[i]
        else:
            character += s[i]
             
    # Sort the string of digits
    Digits = list(digits)
    Digits.sort()
     
    # Sort the string of characters
    Character = list(character)
    Character.sort()
     
    # Stores odd and even characters
    OddChar, EvenChar = "", ""
     
    # Seperate odd and even digits
    for i in range(len(Digits)):
        if ((ord(digits[i]) - ord('0')) % 2 == 1):
            OddChar += Digits[i]
        else:
            EvenChar += Digits[i]
             
    # Concatenate strings in the order
    # odd characters followed by even
    OddChar += EvenChar
    EvenChar = ""
     
    # Seperate the odd and even chars
    for i in range(len(Character)):
        if ((ord(Character[i]) - ord('a')) % 2 == 1):
            OddChar += Character[i]
        else:
            EvenChar += Character[i]
             
    # Final string
    OddChar += EvenChar
     
    # Return the final string
    return OddChar
     
# Driver Code
 
# Given string
s = "abcd1234"
 
# Returns the sorted string
print(sortString(list(s)))
 
# This code is contributed by divyesh072019

C#

// C# program for the above approach
using System;
class GFG
{
 
  // Function to sort the string s
  // based on the given conditions
  static string sortString(char[] s)
  {
 
    // Stores digits of given string
    string digits = "";
 
    // Stores characters of given string
    string character = "";
 
    // Iterate over characters of the string
    for (int i = 0; i < s.Length; ++i)
    {
      if (s[i] >= '0' && s[i] <= '9')
      {
        digits += s[i];
      }
      else
      {
        character += s[i];
      }
    }
 
    // Sort the string of digits
    char[] Digits = digits.ToCharArray();
    Array.Sort(Digits);
 
    // Sort the string of characters
    char[] Character = character.ToCharArray();
    Array.Sort(Character);
 
    // Stores odd and even characters
    string OddChar = "", EvenChar = "";
 
    // Seperate odd and even digits
    for (int i = 0; i < Digits.Length; ++i)
    {
      if ((digits[i] - '0') % 2 == 1)
      {
        OddChar += Digits[i];
      }
      else
      {
        EvenChar += Digits[i];
      }
    }
 
    // Concatenate strings in the order
    // odd characters followed by even
    OddChar += EvenChar;
    EvenChar = "";
 
    // Seperate the odd and even chars
    for (int i = 0; i < Character.Length; ++i)
    {
 
      if ((Character[i] - 'a') % 2 == 1)
      {
        OddChar += Character[i];
      }
      else
      {
        EvenChar += Character[i];
      }
    }
 
    // Final string
    OddChar += EvenChar;
 
    // Return the final string
    return OddChar;
  }
 
  // Driver code
  static void Main()
  {
     
    // Given string
    string s = "abcd1234";
 
    // Returns the sorted string
    Console.WriteLine(sortString(s.ToCharArray()));
  }
}
 
// This code is contributed by divyehrabadiya07

输出：

1324bdac

时间复杂度： O(N)
辅助空间： O(N)