程序查找字符串中字符的ASCII值的XOR

给定字符串str ，任务是查找字符串中字符的ASCII值的XOR。

例子：

Input: str = “Geeks”
Output: 95
ASCII value of G = 71
ASCII value of e = 101
ASCII value of e = 101
ASCII value of k = 107
ASCII value of s = 115
XOR of ASCII values = 71 ^ 101 ^ 101 ^ 107 ^ 115 = 95

Input: str = “GfG”
Output: 102

方法：想法是一个个地找出每个字符的ASCII值，然后找到这些值的XOR值。

下面是上述方法的实现：

C++

// C++ program to find XOR of ASCII
// value of characters in string
  
#include 
using namespace std;
  
// Function to find the XOR of ASCII
// value of characters in string
int XorAscii(string str, int len)
{
  
    // store value of first character
    int ans = int(str[0]);
  
    for (int i = 1; i < len; i++) {
  
        // Traverse string to find the XOR
        ans = (ans ^ (int(str[i])));
    }
  
    // Return the XOR
    return ans;
}
  
// Driver code
int main()
{
  
    string str = "geeksforgeeks";
    int len = str.length();
    cout << XorAscii(str, len) << endl;
  
    str = "GfG";
    len = str.length();
    cout << XorAscii(str, len);
  
    return 0;
}

Java

// Java program to find XOR of ASCII
// value of characters in String
class GFG{
   
// Function to find the XOR of ASCII
// value of characters in String
static int XorAscii(String str, int len)
{
   
    // store value of first character
    int ans = (str.charAt(0));
   
    for (int i = 1; i < len; i++) {
   
        // Traverse String to find the XOR
        ans = (ans ^ ((str.charAt(i))));
    }
   
    // Return the XOR
    return ans;
}
   
// Driver code
public static void main(String[] args)
{
   
    String str = "geeksforgeeks";
    int len = str.length();
    System.out.print(XorAscii(str, len) +"\n");
   
    str = "GfG";
    len = str.length();
    System.out.print(XorAscii(str, len)); 
}
}
  
// This code is contributed by PrinciRaj1992

Python3

# Python3 program to find XOR of ASCII
# value of characters in str1ing
  
# Function to find the XOR of ASCII
# value of characters in str1ing
def XorAscii(str1, len1):
  
    # store value of first character
    ans = ord(str1[0])
  
    for i in range(1,len1):
  
        # Traverse str1ing to find the XOR
        ans = (ans ^ (ord(str1[i])))
  
    # Return the XOR
    return ans
  
# Driver code
str1 = "geeksforgeeks"
len1 = len(str1)
print(XorAscii(str1, len1))
  
str1 = "GfG"
len1 = len(str1)
print(XorAscii(str1, len1))
  
# This code is contributed by mohit kumar 29

C#

// C# program to find XOR of ASCII
// value of characters in String
using System;
  
class GFG{
    
// Function to find the XOR of ASCII
// value of characters in String
static int XorAscii(String str, int len)
{
    
    // store value of first character
    int ans = (str[0]);
    
    for (int i = 1; i < len; i++) {
    
        // Traverse String to find the XOR
        ans = (ans ^ ((str[i])));
    }
    
    // Return the XOR
    return ans;
}
    
// Driver code
public static void Main(String[] args)
{
    
    String str = "geeksforgeeks";
    int len = str.Length;
    Console.Write(XorAscii(str, len) +"\n");
    
    str = "GfG";
    len = str.Length;
    Console.Write(XorAscii(str, len)); 
}
}
  
// This code is contributed by PrinciRaj1992

输出：

123
102

时间复杂度： O(N) ，其中N是字符串的长度。