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。
【查找两个字符串中不常见的字符|S2】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# 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：