从两个可能的值向变量分配其他值

假设变量x只能有两个可能的值a和b，并且您希望将x的值分配给它的当前值。无需使用任何条件运算符即可高效地执行此操作。

注意：我们不允许检查x的当前值。

例子：

Input : a = 10, b = 15, x = a
Output : x = 15
Explanation x = 10, currently x has value of a (which is 10), we need to change it to 15.

Input : a = 9, b = 11, x = b
Output : x = 9

我们可以使用if条件解决此问题，但是我们不允许这样做。

if (x == a) 
   x = b;
else x = a;

我们本可以使用Ternary运算符，它还会检查x的当前值，并根据它分配新值。所以我们也不能使用这种方法

x = x == a ? b : a;

但是，我们不允许检查x的值，因此上述解决方案均无效。

解决方案1：仅使用算术运算运算符，我们可以执行此操作

x = a + b - x

这样，x的内容将在每次执行时在a和b之间交替

C++

// CPP program to change value of x
// according to its current value.
#include 
using namespace std;
  
// Function to alternate the values
void alternate(int& a, int& b, int& x)
{
    x = a + b - x;
}
  
// Main function
int main()
{
    int a = -10;
    int b = 15;
    int x = a; 
    cout << "x is : " << x;
  
    alternate(a, b, x); 
  
    cout << "\nAfter change ";
    cout << "\nx is : " << x;
}

Java

// Java program to change value of x
// according to its current value.
import java.util.*;
  
class solution
{
  
// Function to alternate the values
static void alternate(int a, int b, int x)
{
    x = a + b - x;
    System.out.println("After change"+"\n"+" x is : "+x);
}
  
// Main function
public static void main(String args[])
{
    int a = -10;
    int b = 15;
    int x = a; 
    System.out.println("x is : "+x);
    alternate(a, b, x); 
}
}

Python3

# Python3 program to change value 
# of x according to its current value. 
  
# Function to alternate the values 
def alternate(a,b,x):
    x = a+b-x
    print("After change x is:",x)
  
  
# Driver code
if __name__=='__main__':
    a = -10
    b = 15
    x = a
    print("x is:",x)
    alternate(a,b,x)
  
# This code is contributed by 
# Shrikant13

C#

// C# program to change value of x
// according to its current value.
  
using System;
class gfg
{
 // Function to alternate the values
 public void alternate(ref int a, ref int b, ref int x)
   //'ref' indicates the references 
 {
    x = a + b - x;
 }
}
  
// Main function
class geek
{
 public static int Main()
 {
    gfg g = new gfg();
    int a = -10;
    int b = 15;
    int x = a; 
    Console.WriteLine("x is : {0}" , x);
  
    g.alternate(ref a, ref b, ref x); 
  
    Console.WriteLine ("After change ");
    Console.WriteLine("x is : {0}", x);
    return 0;
 }
}
//This code is contributed by Soumik

输出：

x is : -10
After change 
x is : 15

时间复杂度：此方法的时间复杂度为O(1)
空间复杂度：此方法的空间复杂度为O(1)

解决方案2：一种更好，更有效的方法是使用按位XOR操作。

x = a ^ b ^ x

C++

// CPP program to change value of x
// according to its current value.
#include 
using namespace std;
  
// Function to alternate the values
void alternate(int& a, int& b, int& x)
{
    x = a ^ b ^ x;
}
  
// Main function
int main()
{
    int a = -10;
    int b = 15;
    int x = a;
    cout << "x is : " << x;
  
    alternate(a, b, x); 
  
    cout << "\nAfter exchange ";
    cout << "\nx is : " << x;
      
    return 0;
}

Java

// Java program to change value of x
// according to its current value.
  
class GFG {
// Function to alternate the values
  
    static int alternate(int a, int b, int x) {
        return x = a ^ b ^ x;
    }
  
// Main function
    public static void main(String[] args) {
        int a = -10;
        int b = 15;
        int x = a;
        System.out.print("x is : " + x);
  
        x = alternate(a, b, x);
  
        System.out.print("\nAfter exchange ");
        System.out.print("\nx is : " + x);
  
    }
} 
  
// This code is contributed by 29AjayKumar

Python3

# Python3 program to change value of x
# according to its current value.
  
# Function to alternate the values
def alternate(a, b, x):
    x = a ^ b ^ x
    print("After exchange")
    print("x is", x)
  
# Driver code
a = -10
b = 15
x = a
print("x is", x)
alternate(a, b, x)
  
# This code is contributed 
# by Shrikant13

C＃

// C# program to change value of x 
// according to its current value. 
using System;
public class GFG { 
// Function to alternate the values 
  
    static int alternate(int a, int b, int x) { 
        return x = a ^ b ^ x; 
    } 
  
// Main function 
    public static void Main() { 
        int a = -10; 
        int b = 15; 
        int x = a; 
        Console.Write("x is : " + x); 
  
        x = alternate(a, b, x); 
  
        Console.Write("\nAfter exchange "); 
        Console.Write("\nx is : " + x); 
  
    } 
} 
/*This code is contributed by Rajput-Ji*/

输出：

x is : -10
After exchange 
x is : 15

时间复杂度：此方法的时间复杂度为O(1)
空间复杂度：此方法的空间复杂度为O(1)