查找两个字符串中不常见的字符|S2

本文概述

• C ++
• Java
• C#
• Python3

给定两个字符串str1和str2, 任务是查找并打印两个给定字符串中不常见的字符, 并且不使用多余的空间。此处不常见的字符表示该字符出现在一个字符串中, 或者出现在另一个字符串中, 但都不出现在两个字符串中。字符串仅包含小写字符, 并且可以包含重复项。

例子：

输入：str1 ="characters", str2 ="alphabets"
输出：b c l p r
输入：str1 ="geeksforgeeks", str2 ="geeksquiz"
输出：f i o q r u z

方法:这里讨论了一种使用散列的方法。这个问题也可以通过使用位操作来解决。

该方法使用2个变量, 分别存储每个字符的ASCII码左移1的按位或-97, 即0表示" a", 1表示" b", 依此类推。对于这两个字符串, 在执行这些按位运算后, 我们都会得到一个整数。现在, 这两个整数的异或运算将仅在表示罕见字符的位置将二进制位设为1。打印这些位置的字符值。

下面是上述方法的实现：

C ++

//C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
//Function to print the uncommon
//characters in the given string
//in sorted order
void printUncommon(string str1, string str2)
{
     int a1 = 0, a2 = 0;
     for ( int i = 0; i <str1.length(); i++) {
  
         //Converting character to ASCII code
         int ch = int (str1[i]) - 'a' ;
  
         //Bit operation
         a1 = a1 | (1 <<ch);
     }
     for ( int i = 0; i <str2.length(); i++) {
  
         //Converting character to ASCII code
         int ch = int (str2[i]) - 'a' ;
  
         //Bit operation
         a2 = a2 | (1 <<ch);
     }
  
     //XOR operation leaves only uncommon
     //characters in the ans variable
     int ans = a1 ^ a2;
  
     int i = 0;
     while (i <26) {
         if (ans % 2 == 1) {
             cout <<char ( 'a' + i);
         }
         ans = ans /2;
         i++;
     }
}
  
//Driver code
int main()
{
     string str1 = "lsbin" ;
     string str2 = "geeksquiz" ;
  
     printUncommon(str1, str2);
  
     return 0;
}

Java

//Java implementation of the approach
class GFG 
{
  
     //Function to print the uncommon
     //characters in the given string
     //in sorted order
     static void printUncommon(String str1, String str2) 
     {
         int a1 = 0 , a2 = 0 ;
         for ( int i = 0 ; i <str1.length(); i++)
         {
  
             //Converting character to ASCII code
             int ch = (str1.charAt(i)) - 'a' ;
  
             //Bit operation
             a1 = a1 | ( 1 <<ch);
         }
         for ( int i = 0 ; i <str2.length(); i++) 
         {
  
             //Converting character to ASCII code
             int ch = (str2.charAt(i)) - 'a' ;
  
             //Bit operation
             a2 = a2 | ( 1 <<ch);
         }
  
         //XOR operation leaves only uncommon
         //characters in the ans variable
         int ans = a1 ^ a2;
  
         int i = 0 ;
         while (i <26 ) 
         {
             if (ans % 2 == 1 )
             {
                 System.out.print(( char ) ( 'a' + i));
             }
             ans = ans /2 ;
             i++;
         }
     }
  
     //Driver code
     public static void main(String[] args)
     {
         String str1 = "lsbin" ;
         String str2 = "geeksquiz" ;
  
         printUncommon(str1, str2);
     }
}
  
//This code contributed by Rajput-Ji

C#

//C# implementation of the approach
using System;
  
class GFG
{
      
//Function to print the uncommon
//characters in the given string
//in sorted order
static void printUncommon( string str1, string str2)
{
     int a1 = 0, a2 = 0;
     for ( int i = 0; i <str1.Length; i++)
     {
  
         //Converting character to ASCII code
         int ch = (str1[i] - 'a' );
  
         //Bit operation
         a1 = a1 | (1 <<ch);
     }
     for ( int i = 0; i <str2.Length; i++) 
     {
  
         //Converting character to ASCII code
         int ch = (str2[i] - 'a' );
  
         //Bit operation
         a2 = a2 | (1 <<ch);
     }
  
     //XOR operation leaves only uncommon
     //characters in the ans variable
     int ans = a1 ^ a2;
  
     int j = 0;
     while (j <26) 
     {
         if (ans % 2 == 1) 
         {
             Console.Write(( char )( 'a' + j));
         }
         ans = ans /2;
         j++;
     }
}
  
//Driver code
public static void Main()
{
     string str1 = "lsbin" ;
     string str2 = "geeksquiz" ;
  
     printUncommon(str1, str2);
  
}
}
  
//This code is contributed by SoM15242

Python3

# Python3 implementation of the approach 
  
# Function to print the uncommon 
# characters in the given string 
# in sorted order 
def printUncommon(str1, str2) : 
  
     a1 = 0 ; a2 = 0 ; 
      
     for i in range ( len (str1)) :
  
         # Converting character to ASCII code 
         ch = ord (str1[i]) - ord ( 'a' ); 
  
         # Bit operation 
         a1 = a1 | ( 1 <<ch); 
      
     for i in range ( len (str2)) : 
  
         # Converting character to ASCII code 
         ch = ord (str2[i]) - ord ( 'a' ); 
  
         # Bit operation 
         a2 = a2 | ( 1 <<ch); 
  
     # XOR operation leaves only uncommon 
     # characters in the ans variable 
     ans = a1 ^ a2; 
  
     i = 0 ; 
     while (i <26 ) :
         if (ans % 2 = = 1 ) :
             print ( chr ( ord ( 'a' ) + i), end = ""); 
          
         ans = ans //2 ; 
         i + = 1 ; 
  
# Driver code 
if __name__ = = "__main__" : 
  
     str1 = "lsbin" ; 
     str2 = "geeksquiz" ; 
  
     printUncommon(str1, str2); 
      
     # This code is contributed by AnkitRai01

输出如下：

fioqruz

---