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📜  检查两个数字是否是彼此的位旋转

📅  最后修改于: 2021-05-25 02:49:47             🧑  作者: Mango

给定两个正整数x和y,请检查是否通过旋转另一个整数来获得一个整数。

Input constraint: 0 < x, y < 2^32 

位旋转:旋转(或循环移位)是与移位类似的操作,不同之处在于,一端掉落的位放回另一端。
有关位旋转的更多信息,请参见此处
范例1:

Input : a = 8, b = 1
Output : yes

Explanation :
Represntation of a = 8 : 0000 0000 0000 0000 0000 0000 0000 1000
Represntation of b = 1 : 0000 0000 0000 0000 0000 0000 0000 0001
If we rotate a by 3 units right we get b, hence answer is yes

范例2:

Input : a = 122, b = 2147483678
Output : yes

Explanation :
Represntation of a = 122        : 0000 0000 0000 0000 0000 0000 0111 1010
Represntation of b = 2147483678 : 1000 0000 0000 0000 0000 0000 0001 1110
If we rotate a by 2 units right we get b, hence answer is yes

由于x,y> 0且x,y <2 ^ 32,因此可以表示x或y的总位数为32。
因此,我们需要找到x的所有32种可能的旋转并将其与y进行比较,直到x和y不相等。
为此,我们使用64位临时变量x64,这是x与x串联的结果。
x64的前32位与x的位相同,后32位也与x64的位相同。
然后,我们继续在右侧将x64移位1,并将x64最右边的32位与y进行比较。
这样,由于旋转,我们将能够获得所有可能的位组合。
这是上述算法的实现。

C++
// C++ program to check if two numbers are bit rotations
// of each other.
#include 
using namespace std;
 
// function to check if  two numbers are equal
// after bit rotation
bool isRotation(unsigned int x, unsigned int y)
{
    // x64 has concatenation of x with itself.
    unsigned long long int x64 = x | ((unsigned long long int)x << 32);
 
    while (x64 >= y)
    {
        // comapring only last 32 bits
        if (unsigned(x64) == y)
            return true;
 
        // right shift by 1 unit
        x64 >>= 1;
    }
    return false;
}
 
// driver code to test above function
int main()
{
    unsigned int x = 122;
    unsigned int y = 2147483678;
 
    if (isRotation(x, y))
        cout << "yes" << endl;
    else
        cout << "no" << endl;
 
    return 0;
}


Java
// Java program to check if two numbers are bit rotations
// of each other.
class GFG {
 
// function to check if two numbers are equal
// after bit rotation
    static boolean isRotation(long x, long y) {
        // x64 has concatenation of x with itself.
        long x64 = x | (x << 32);
 
        while (x64 >= y) {
            // comapring only last 32 bits
            if (x64 == y) {
                return true;
            }
 
            // right shift by 1 unit
            x64 >>= 1;
        }
        return false;
    }
 
// driver code to test above function
    public static void main(String[] args) {
        long x = 122;
        long y = 2147483678L;
 
        if (isRotation(x, y) == false) {
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }
    }
}
 
// This code is contributed by 29AjayKumar


Python3
# Python3 program to check if two
# numbers are bit rotations of each other.
 
# function to check if two numbers
# are equal after bit rotation
def isRotation(x, y) :
     
    # x64 has concatenation of x
    # with itself.
    x64 = x | (x << 32)
     
    while (x64 >= y) :
         
        # comapring only last 32 bits
        if ((x64) == y) :
            return True
 
        # right shift by 1 unit
        x64 >>= 1
 
    return False
 
# Driver Code
if __name__ == "__main__" :
 
    x = 122
    y = 2147483678
     
    if (isRotation(x, y) == False) :
        print("yes")
    else :
        print("no")
 
# This code is contributed by Ryuga


C#
// C# program to check if two numbers
// are bit rotations of each other.
using System;
 
class GFG
{
 
// function to check if two numbers
// are equal after bit rotation
static bool isRotation(long x, long y)
{
    // x64 has concatenation of
    // x with itself.
    long x64 = x | (x << 32);
 
    while (x64 >= y)
    {
        // comapring only last 32 bits
        if (x64 == y)
        {
            return true;
        }
 
        // right shift by 1 unit
        x64 >>= 1;
    }
    return false;
}
 
// Driver Code
public static void Main()
{
    long x = 122;
    long y = 2147483678L;
 
    if (isRotation(x, y) == false)
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
}
 
// This code is contributed
// by 29AjayKumar


PHP
= $y)
    {
        // comapring only last 32 bits
        if (($x64) == $y)
            return 1;
 
        // right shift by 1 unit
        $x64 >>= 1;
    }
    return -1;
}
 
// Driver Code
$x = 122;
$y = 2147483678;
 
if (isRotation($x, $y))
    echo "yes" ,"\n";
else
    echo "no" ,"\n";
 
// This code is contributed by aj_36
?>


Javascript


输出 :

yes