旋转数字位的 Java 程序
比特旋转:旋转(或循环移位)是一种类似移位的操作，只是一端脱落的比特放回另一端。 左旋转时，左端脱落的比特放回右端。 右旋转时，右端脱落的钻头放回左端。
示例: 让 n 用 8 位存储。n = 11100101 向左旋转 3，则 n = 00101111(左移 3 位，前 3 位放回最后)。如果使用 16 位或 32 位存储 n，那么 n (000…11100101)的左旋转变为 00..00 11100101 000。 如果使用 8 位存储 n，n = 11100101 向右旋转 3，则 n = 10111100(右移 3，最后 3 位放回第一位)。如果使用 16 位或 32 位存储 n，那么 n (000…11100101)向右旋转 3 就变成了 101 000..00 11100 。
Java 语言(一种计算机语言，尤用于创建网站)
// Java code to rotate bits
// of number
class GFG
{
static final int INT_BITS = 32;
/*Function to left rotate n by d bits*/
static int leftRotate(int n, int d) {
/* In n<<d, last d bits are 0.
To put first 3 bits of n at
last, do bitwise or of n<<d with
n >>(INT_BITS - d) */
return (n << d) | (n >> (INT_BITS - d));
}
/*Function to right rotate n by d bits*/
static int rightRotate(int n, int d) {
/* In n>>d, first d bits are 0.
To put last 3 bits of at
first, do bitwise or of n>>d
with n <<(INT_BITS - d) */
return (n >> d) | (n << (INT_BITS - d));
}
// Driver code
public static void main(String arg[])
{
int n = 16;
int d = 2;
System.out.print("Left Rotation of " + n +
" by " + d + " is ");
System.out.print(leftRotate(n, d));
System.out.print("
Right Rotation of " + n +
" by " + d + " is ");
System.out.print(rightRotate(n, d));
}
}
// This code is contributed by Anant Agarwal.
输出:
Left Rotation of 16 by 2 is 64
Right Rotation of 16 by 2 is 4
16 位数字:
Java 语言(一种计算机语言，尤用于创建网站)
/*package whatever //do not write package name here */
import java.io.*;
class GFG {
public static void main (String[] args) {
int N=28;
int D=2;
rotate(N,D);
}
static void rotate(int N, int D)
{
// your code here
int t=16;
int left= ((N<<D) | N>>(t-D)) & 0xFFFF;
int right=((N>>D) | N<<(t-D)) & 0xFFFF;
System.out.println(left);
System.out.println(right);
}
}
Output:
Left Rotation of 28 by 2 is 112
Right Rotation of 28 by 2 is 7
时间复杂度 : O(1)
空间复杂度 : O(1)
更多详情请参考旋转数字位整篇文章！
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