用于查找字符串中字符的 ASCII 值的异或的程序

给定一个字符串str ，任务是找到字符串中字符的 ASCII 值的异或。
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 
 

做法：思路是将每个字符的ASCII值一一找出，并求出这些值的异或值。
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

Javascript

 
// Javascript program to find XOR of ASCII
// value of characters in string
 
// Function to find the XOR of ASCII
// value of characters in string
function XorAscii(str, len)
{
 
    // store value of first character
    let ans = str.codePointAt(0);
 
    for (let i = 1; i < len; i++) {
 
        // Traverse string to find the XOR
        ans = (ans ^ (str.codePointAt(i)));
    }
 
    // Return the XOR
    return ans;
}
 
// Driver code
 
    let str = "geeksforgeeks";
    let len = str.length;
    document.write(XorAscii(str, len) + "<br>");
 
    str = "GfG";
    len = str.length;
    document.write(XorAscii(str, len));
 

输出：

123
102

时间复杂度： O(N) ，其中 N 是字符串的长度。

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