在某些分支机构价格昂贵的稀有机器上,以下明显的最小值查找方法可能会很慢,因为它使用分支机构。
c
/* The obvious approach to find minimum (involves branching) */
int min(int x, int y)
{
return (x < y) ? x : y
}
C++
// C++ program to Compute the minimum
// or maximum of two integers without
// branching
#include
using namespace std;
class gfg
{
/*Function to find minimum of x and y*/
public:
int min(int x, int y)
{
return y ^ ((x ^ y) & -(x < y));
}
/*Function to find maximum of x and y*/
int max(int x, int y)
{
return x ^ ((x ^ y) & -(x < y));
}
};
/* Driver code */
int main()
{
gfg g;
int x = 15;
int y = 6;
cout << "Minimum of " << x <<
" and " << y << " is ";
cout << g. min(x, y);
cout << "\nMaximum of " << x <<
" and " << y << " is ";
cout << g.max(x, y);
getchar();
}
// This code is contributed by SoM15242
C
// C program to Compute the minimum
// or maximum of two integers without
// branching
#include
/*Function to find minimum of x and y*/
int min(int x, int y)
{
return y ^ ((x ^ y) & -(x < y));
}
/*Function to find maximum of x and y*/
int max(int x, int y)
{
return x ^ ((x ^ y) & -(x < y));
}
/* Driver program to test above functions */
int main()
{
int x = 15;
int y = 6;
printf("Minimum of %d and %d is ", x, y);
printf("%d", min(x, y));
printf("\nMaximum of %d and %d is ", x, y);
printf("%d", max(x, y));
getchar();
}
Java
// Java program to Compute the minimum
// or maximum of two integers without
// branching
public class AWS {
/*Function to find minimum of x and y*/
static int min(int x, int y)
{
return y ^ ((x ^ y) & -(x << y));
}
/*Function to find maximum of x and y*/
static int max(int x, int y)
{
return x ^ ((x ^ y) & -(x << y));
}
/* Driver program to test above functions */
public static void main(String[] args) {
int x = 15;
int y = 6;
System.out.print("Minimum of "+x+" and "+y+" is ");
System.out.println(min(x, y));
System.out.print("Maximum of "+x+" and "+y+" is ");
System.out.println( max(x, y));
}
}
Python3
# Python3 program to Compute the minimum
# or maximum of two integers without
# branching
# Function to find minimum of x and y
def min(x, y):
return y ^ ((x ^ y) & -(x < y))
# Function to find maximum of x and y
def max(x, y):
return x ^ ((x ^ y) & -(x < y))
# Driver program to test above functions
x = 15
y = 6
print("Minimum of", x, "and", y, "is", end=" ")
print(min(x, y))
print("Maximum of", x, "and", y, "is", end=" ")
print(max(x, y))
# This code is contributed
# by Smitha Dinesh Semwal
C#
using System;
// C# program to Compute the minimum
// or maximum of two integers without
// branching
public class AWS
{
/*Function to find minimum of x and y*/
public static int min(int x, int y)
{
return y ^ ((x ^ y) & -(x << y));
}
/*Function to find maximum of x and y*/
public static int max(int x, int y)
{
return x ^ ((x ^ y) & -(x << y));
}
/* Driver program to test above functions */
public static void Main(string[] args)
{
int x = 15;
int y = 6;
Console.Write("Minimum of " + x + " and " + y + " is ");
Console.WriteLine(min(x, y));
Console.Write("Maximum of " + x + " and " + y + " is ");
Console.WriteLine(max(x, y));
}
}
// This code is contributed by Shrikant13
PHP
Javascript
C++
#include
using namespace std;
#define CHARBIT 8
/*Function to find minimum of x and y*/
int min(int x, int y)
{
return y + ((x - y) & ((x - y) >>
(sizeof(int) * CHARBIT - 1)));
}
/*Function to find maximum of x and y*/
int max(int x, int y)
{
return x - ((x - y) & ((x - y) >>
(sizeof(int) * CHARBIT - 1)));
}
/* Driver code */
int main()
{
int x = 15;
int y = 6;
cout<<"Minimum of "<
C
#include
#define CHAR_BIT 8
/*Function to find minimum of x and y*/
int min(int x, int y)
{
return y + ((x - y) & ((x - y) >>
(sizeof(int) * CHAR_BIT - 1)));
}
/*Function to find maximum of x and y*/
int max(int x, int y)
{
return x - ((x - y) & ((x - y) >>
(sizeof(int) * CHAR_BIT - 1)));
}
/* Driver program to test above functions */
int main()
{
int x = 15;
int y = 6;
printf("Minimum of %d and %d is ", x, y);
printf("%d", min(x, y));
printf("\nMaximum of %d and %d is ", x, y);
printf("%d", max(x, y));
getchar();
}
Java
// JAVA implementation of above approach
class GFG
{
static int CHAR_BIT = 4;
static int INT_BIT = 8;
/*Function to find minimum of x and y*/
static int min(int x, int y)
{
return y + ((x - y) & ((x - y) >>
(INT_BIT * CHAR_BIT - 1)));
}
/*Function to find maximum of x and y*/
static int max(int x, int y)
{
return x - ((x - y) & ((x - y) >>
(INT_BIT * CHAR_BIT - 1)));
}
/* Driver code */
public static void main(String[] args)
{
int x = 15;
int y = 6;
System.out.println("Minimum of "+x+" and "+y+" is "+min(x, y));
System.out.println("Maximum of "+x+" and "+y+" is "+max(x, y));
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 implementation of the approach
import sys;
CHAR_BIT = 8;
INT_BIT = sys.getsizeof(int());
#Function to find minimum of x and y
def Min(x, y):
return y + ((x - y) & ((x - y) >>
(INT_BIT * CHAR_BIT - 1)));
#Function to find maximum of x and y
def Max(x, y):
return x - ((x - y) & ((x - y) >>
(INT_BIT * CHAR_BIT - 1)));
# Driver code
x = 15;
y = 6;
print("Minimum of", x, "and",
y, "is", Min(x, y));
print("Maximum of", x, "and",
y, "is", Max(x, y));
# This code is contributed by PrinciRaj1992
C#
// C# implementation of above approach
using System;
class GFG
{
static int CHAR_BIT = 8;
/*Function to find minimum of x and y*/
static int min(int x, int y)
{
return y + ((x - y) & ((x - y) >>
(sizeof(int) * CHAR_BIT - 1)));
}
/*Function to find maximum of x and y*/
static int max(int x, int y)
{
return x - ((x - y) & ((x - y) >>
(sizeof(int) * CHAR_BIT - 1)));
}
/* Driver code */
static void Main()
{
int x = 15;
int y = 6;
Console.WriteLine("Minimum of "+x+" and "+y+" is "+min(x, y));
Console.WriteLine("Maximum of "+x+" and "+y+" is "+max(x, y));
}
}
// This code is contributed by mits
Javascript
Java
/*package whatever //do not write package name here */
import java.io.*;
class GFG {
public static void main(String []args){
System.out.println( max(2,3) ); //3
System.out.println( max(2,-3) ); //2
System.out.println( max(-2,-3) ); //-2
System.out.println( min(2,3) ); //2
System.out.println( min(2,-3) ); //-3
System.out.println( min(-2,-3) ); //-3
}
public static int max(int x, int y){
int abs = absbit32(x,y);
return (x + y + abs)/2;
}
public static int min(int x, int y){
int abs = absbit32(x,y);
return (x + y - abs)/2;
}
public static int absbit32(int x, int y){
int sub = x - y;
int mask = (sub >> 31);
return (sub ^ mask) - mask;
}
}
Python3
def max(x, y):
abs = absbit32(x,y)
return (x + y + abs)//2
def min(x, y):
abs = absbit32(x,y)
return (x + y - abs)//2
def absbit32( x, y):
sub = x - y
mask = (sub >> 31)
return (sub ^ mask) - mask
# Driver code
print( max(2,3) ) #3
print( max(2,-3) ) #2
print( max(-2,-3) ) #-2
print( min(2,3) ) #2
print( min(2,-3) ) #-3
print( min(-2,-3) ) #-3
# This code is contributed by rohitsingh07052.
C#
using System;
class GFG{
public static void Main(String []args)
{
Console.WriteLine(max(2, 3)); //3
Console.WriteLine(max(2, -3)); //2
Console.WriteLine(max(-2, -3)); //-2
Console.WriteLine(min(2, 3)); //2
Console.WriteLine(min(2, -3)); //-3
Console.WriteLine(min(-2, -3)); //-3
}
public static int max(int x, int y)
{
int abs = absbit32(x, y);
return (x + y + abs) / 2;
}
public static int min(int x, int y)
{
int abs = absbit32(x, y);
return (x + y - abs) / 2;
}
public static int absbit32(int x, int y)
{
int sub = x - y;
int mask = (sub >> 31);
return (sub ^ mask) - mask;
}
}
// This code is contributed by Amit Katiyar
Javascript
以下是在不使用分支的情况下获得最小值(或最大值)的方法。通常,显而易见的方法是最好的。
方法1(使用XOR和比较运算符)
x和y的最小值将是
y ^ ((x ^ y) & -(x < y))
之所以起作用是因为如果x 如果x> y,则-(x 在某些机器上,将(x 输出: 方法2(使用减法和移位) ,那么我们可以使用以下方法,它们更快,因为(x – y)只需要被评估一次。 此方法将x和y的减法偏移31(如果整数的大小为32)。如果(xy)小于0,则(x -y)>> 31将为1。如果(xy)大于或等于0,则(x -y)>> 31将为0。 请注意,1989年的ANSI C规范未指定有符号右移的结果,因此上述方法不可移植。如果在溢出时引发异常,则x和y的值应为无符号或强制转换为无符号,以免不必要地抛出异常,但是右移需要带符号的操作数在负数时产生所有一位,因此强制转换在那里签名。 方法3(使用绝对值) 查找具有绝对值的最大/最小值的通用公式是: 找到的最小数字是: 因此,如果我们可以使用按位运算来找到绝对值,则可以在不使用if条件的情况下找到最大值/最小值。在这里可以找到按位运算的绝对方法: 步骤1)将掩码设置为整数右移31(假设整数存储为2的补码32位值,并且右移运算符符号扩展)。 步骤2)将遮罩与数字进行XOR 步骤3)从步骤2的结果中减去掩码并返回结果。 因此,我们可以得出以下结论: 来源:x ^ ((x ^ y) & -(x < y));
C++
// C++ program to Compute the minimum
// or maximum of two integers without
// branching
#include
C
// C program to Compute the minimum
// or maximum of two integers without
// branching
#include
Java
// Java program to Compute the minimum
// or maximum of two integers without
// branching
public class AWS {
/*Function to find minimum of x and y*/
static int min(int x, int y)
{
return y ^ ((x ^ y) & -(x << y));
}
/*Function to find maximum of x and y*/
static int max(int x, int y)
{
return x ^ ((x ^ y) & -(x << y));
}
/* Driver program to test above functions */
public static void main(String[] args) {
int x = 15;
int y = 6;
System.out.print("Minimum of "+x+" and "+y+" is ");
System.out.println(min(x, y));
System.out.print("Maximum of "+x+" and "+y+" is ");
System.out.println( max(x, y));
}
}
Python3
# Python3 program to Compute the minimum
# or maximum of two integers without
# branching
# Function to find minimum of x and y
def min(x, y):
return y ^ ((x ^ y) & -(x < y))
# Function to find maximum of x and y
def max(x, y):
return x ^ ((x ^ y) & -(x < y))
# Driver program to test above functions
x = 15
y = 6
print("Minimum of", x, "and", y, "is", end=" ")
print(min(x, y))
print("Maximum of", x, "and", y, "is", end=" ")
print(max(x, y))
# This code is contributed
# by Smitha Dinesh Semwal
C#
using System;
// C# program to Compute the minimum
// or maximum of two integers without
// branching
public class AWS
{
/*Function to find minimum of x and y*/
public static int min(int x, int y)
{
return y ^ ((x ^ y) & -(x << y));
}
/*Function to find maximum of x and y*/
public static int max(int x, int y)
{
return x ^ ((x ^ y) & -(x << y));
}
/* Driver program to test above functions */
public static void Main(string[] args)
{
int x = 15;
int y = 6;
Console.Write("Minimum of " + x + " and " + y + " is ");
Console.WriteLine(min(x, y));
Console.Write("Maximum of " + x + " and " + y + " is ");
Console.WriteLine(max(x, y));
}
}
// This code is contributed by Shrikant13
的PHP
Java脚本
Minimum of 15 and 6 is 6
Maximum of 15 and 6 is 15
如果我们知道INT_MIN <= (x - y) <= INT_MAX
x和y的最小值将是y + ((x - y) & ((x - y) >>(sizeof(int) * CHAR_BIT - 1)))
因此,如果x> = y,我们得到的最小值为y +(xy)&0,即y。
如果x x - ((x - y) & ((x - y) >> (sizeof(int) * CHAR_BIT - 1)))
C++
#include
C
#include
Java
// JAVA implementation of above approach
class GFG
{
static int CHAR_BIT = 4;
static int INT_BIT = 8;
/*Function to find minimum of x and y*/
static int min(int x, int y)
{
return y + ((x - y) & ((x - y) >>
(INT_BIT * CHAR_BIT - 1)));
}
/*Function to find maximum of x and y*/
static int max(int x, int y)
{
return x - ((x - y) & ((x - y) >>
(INT_BIT * CHAR_BIT - 1)));
}
/* Driver code */
public static void main(String[] args)
{
int x = 15;
int y = 6;
System.out.println("Minimum of "+x+" and "+y+" is "+min(x, y));
System.out.println("Maximum of "+x+" and "+y+" is "+max(x, y));
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 implementation of the approach
import sys;
CHAR_BIT = 8;
INT_BIT = sys.getsizeof(int());
#Function to find minimum of x and y
def Min(x, y):
return y + ((x - y) & ((x - y) >>
(INT_BIT * CHAR_BIT - 1)));
#Function to find maximum of x and y
def Max(x, y):
return x - ((x - y) & ((x - y) >>
(INT_BIT * CHAR_BIT - 1)));
# Driver code
x = 15;
y = 6;
print("Minimum of", x, "and",
y, "is", Min(x, y));
print("Maximum of", x, "and",
y, "is", Max(x, y));
# This code is contributed by PrinciRaj1992
C#
// C# implementation of above approach
using System;
class GFG
{
static int CHAR_BIT = 8;
/*Function to find minimum of x and y*/
static int min(int x, int y)
{
return y + ((x - y) & ((x - y) >>
(sizeof(int) * CHAR_BIT - 1)));
}
/*Function to find maximum of x and y*/
static int max(int x, int y)
{
return x - ((x - y) & ((x - y) >>
(sizeof(int) * CHAR_BIT - 1)));
}
/* Driver code */
static void Main()
{
int x = 15;
int y = 6;
Console.WriteLine("Minimum of "+x+" and "+y+" is "+min(x, y));
Console.WriteLine("Maximum of "+x+" and "+y+" is "+max(x, y));
}
}
// This code is contributed by mits
Java脚本
(x + y + ABS(x-y) )/2
(x + y - ABS(x-y) )/2
mask = n>>31
mask ^ n
(mask^n) - mask
Java
/*package whatever //do not write package name here */
import java.io.*;
class GFG {
public static void main(String []args){
System.out.println( max(2,3) ); //3
System.out.println( max(2,-3) ); //2
System.out.println( max(-2,-3) ); //-2
System.out.println( min(2,3) ); //2
System.out.println( min(2,-3) ); //-3
System.out.println( min(-2,-3) ); //-3
}
public static int max(int x, int y){
int abs = absbit32(x,y);
return (x + y + abs)/2;
}
public static int min(int x, int y){
int abs = absbit32(x,y);
return (x + y - abs)/2;
}
public static int absbit32(int x, int y){
int sub = x - y;
int mask = (sub >> 31);
return (sub ^ mask) - mask;
}
}
Python3
def max(x, y):
abs = absbit32(x,y)
return (x + y + abs)//2
def min(x, y):
abs = absbit32(x,y)
return (x + y - abs)//2
def absbit32( x, y):
sub = x - y
mask = (sub >> 31)
return (sub ^ mask) - mask
# Driver code
print( max(2,3) ) #3
print( max(2,-3) ) #2
print( max(-2,-3) ) #-2
print( min(2,3) ) #2
print( min(2,-3) ) #-3
print( min(-2,-3) ) #-3
# This code is contributed by rohitsingh07052.
C#
using System;
class GFG{
public static void Main(String []args)
{
Console.WriteLine(max(2, 3)); //3
Console.WriteLine(max(2, -3)); //2
Console.WriteLine(max(-2, -3)); //-2
Console.WriteLine(min(2, 3)); //2
Console.WriteLine(min(2, -3)); //-3
Console.WriteLine(min(-2, -3)); //-3
}
public static int max(int x, int y)
{
int abs = absbit32(x, y);
return (x + y + abs) / 2;
}
public static int min(int x, int y)
{
int abs = absbit32(x, y);
return (x + y - abs) / 2;
}
public static int absbit32(int x, int y)
{
int sub = x - y;
int mask = (sub >> 31);
return (sub ^ mask) - mask;
}
}
// This code is contributed by Amit Katiyar
Java脚本
http://graphics.stanford.edu/~seander/bithacks.html#IntegerMinOrMax