检查字符串是否可以重新排列以形成特殊回文

本文概述

• C ++
• Java
• Python3
• C#

给定一个字符串str，任务是检查它是否可以重新排列以得到一个特殊的回文字符串。如果我们能让它打印YES，否则打印NO。

一个字符串被称为特殊回文，它包含同一个字符在回文位置上的一个大写字母和一个小写字母。例如:ABCcba是一个特殊的回文，但ABCcba不是一个特殊的回文。

例子：

Input: str = "ABCcba"
Output: YES

Input: str = "ABCCBA"
Output: NO

方法：检查一个字符的大写字母是否与该字符的小写字母相同。而且应该只出现一个奇特的字符。我们将每个大写字符的频率增加1, 并将每个小写字符的频率减少1。此后, 频率数组中应该为零, 1或-1。如果发生其他情况, 则直接说不, 否则打印是。

下面是上述方法的实现：

C ++

//C++ implementation of the above approach 
#include<bits/stdc++.h>
using namespace std;
  
//Driver code
int main()
{
     string s = "ABCdcba" ;
          
     //creating a array which stores the 
     //frequency of each character
     int u[26] = {0};
     int n = s.length() ;
          
     for ( int i = 0; i <n ; i++)
     {
         //Checking if a character is uppercase or not
         if ( isupper (s[i])) 
         {
             //Increasing by 1 if uppercase
             u展开 - 65] += 1;
         }
         else
         {
             //Decreasing by 1 if lower case 
             u展开 - 97] -= 1 ;
         }
              
     }
     bool f1 = true ;
      
     //Storing the sum of positive 
     //numbers in the frequency array 
     int po = 0 ;
          
     //Storing the sum of negative 
     //numbers in the frequency array
     int ne = 0 ;
          
     for ( int i = 0 ; i <26 ; i++)
     {
         if (u[i]> 0)
             po += u[i] ;
              
         if (u[i] <0)
             ne += u[i] ;
     }
      
     //If all character balances out then its Yes 
     if (po == 0 && ne == 0)
         cout <<( "YES" ) <<endl;
      
     //If there is only 1 character which 
     //does not balances then also it is Yes 
     else if (po == 1 && ne == 0)
         cout <<( "YES" ) <<endl;
          
     else if (po == 0 && ne == -1)
         cout <<( "YES" ) <<endl;
          
     else
         cout <<( "NO" ) <<endl;
          
}
  
//This code is contributed by 
//Surendra_Gangwar

Java

//Java implementation of the above approach 
public class Improve {
  
  
     public static void main(String args[])
     {
         String s = "ABCdcba" ;
          
         //creating a array which stores the  
         //frequency of each character
         int u[] = new int [ 26 ];
         int n = s.length() ;
          
         for ( int i = 0 ; i <n ; i++)
         {
             //Checking if a character is uppercase or not
             if (Character.isUpperCase(s.charAt(i))) 
             {
                 //Increasing by 1 if uppercase
                 u展开 += 1 ;
             }
             else
             {
                 //Decreasing by 1 if lower case 
                 u展开 -= 1 ;
             }
              
         }
         boolean f1 = true ;
          
         //Storing the sum of positive  
         //numbers in the frequency array 
         int po = 0 ;
          
         //Storing the sum of negative  
         //numbers in the frequency array
         int ne = 0 ;
          
         for ( int i = 0 ; i <26 ; i++)
         {
             if (u[i]> 0 )
                 po += u[i] ;
              
             if (u[i] <0 )
                 ne += u[i] ;
         }
          
         //If all character balances out then its Yes 
         if (po == 0 && ne == 0 )
             System.out.println( "YES" ) ;
          
         //If there is only 1 character which  
         //does not balances then also it is Yes 
         else if (po == 1 && ne == 0 )
             System.out.println( "YES" ) ;
          
         else if (po == 0 && ne == - 1 )
             System.out.println( "YES" ) ;
          
         else
             System.out.println( "NO" ) ;
          
          
     }
     //This code is contributed by ANKITRAI1
}

Python3

# Python implementation of the above approach
s = "ABCdcba"
  
# creating a list which stores the 
# frequency of each character
u = [ 0 ] * 26  
n = len (s)
for i in range (n):
     # Checking if a character is uppercase or not
     if (s[i].isupper()):  
         # Increasing by 1 if uppercase
         u[ ord (s[i]) - 65 ] + = 1  
     else :
         # Decreasing by 1 if lower case
         u[ ord (s[i]) - 97 ] - = 1  
fl = True
  
# Storing the sum of positive 
# numbers in the frequency array
po = 0  
  
# Storing the sum of negative 
# numbers in the frequency array
ne = 0  
for i in range ( 26 ):
     if (u[i]> 0 ):
         po + = u[i]
     if (u[i] <0 ):
         ne + = u[i]
  
# If all character balances out then its Yes
if (po = = 0 and ne = = 0 ):  
     print ( "YES" )
  
# If there is only 1 character which 
# does not balances then also it is Yes
elif (po = = 1 and ne = = 0 ):  
     print ( "YES" )
elif (po = = 0 and ne = = - 1 ):
     print ( "YES" )
else :
     print ( "NO" )

C#

//C# implementation of the 
//above approach
using System; 
  
class GFG 
{
public static void Main()
{
     string s = "ABCdcba" ;
      
     //creating a array which stores 
     //the frequency of each character
     int [] u = new int [26];
     int n = s.Length ;
      
     for ( int i = 0; i <n ; i++)
     {
         //Checking if a character is
         //uppercase or not
         if (Char.IsUpper(s[i])) 
         {
             //Increasing by 1 if uppercase
             u展开 - 65] += 1 ;
         }
         else
         {
             //Decreasing by 1 if lower case 
             u展开 - 97] -= 1 ;
         }
     }
  
     //Storing the sum of positive 
     //numbers in the frequency array 
     int po = 0 ;
      
     //Storing the sum of negative 
     //numbers in the frequency array
     int ne = 0 ;
      
     for ( int i = 0 ; i <26 ; i++)
     {
         if (u[i]> 0)
             po += u[i] ;
          
         if (u[i] <0)
             ne += u[i] ;
     }
      
     //If all character balances 
     //out then its Yes 
     if (po == 0 && ne == 0)
         Console.Write( "YES" + "\n" ) ;
      
     //If there is only 1 character which 
     //does not balances then also it is Yes 
     else if (po == 1 && ne == 0)
         Console.Write( "YES" + "\n" ) ;
      
     else if (po == 0 && ne == -1)
         Console.Write( "YES" + "\n" ) ;
      
     else
         Console.Write( "NO" + "\n" ) ;
}
}
  
//This code is contributed 
//by ChitraNayal

输出如下：

YES

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