在本文中，我们将讨论交换给定数字中的每个交替位的代码解决方案，并返回结果数字。我们将使用位操作的概念来解决这个问题，以在不使用任何循环的情况下以恒定时间解决问题。
Problem statement − We are given a number n, we have to swap the pair of bits that are adjacent to each other.
In other words, we have to swap every odd placed bit with its adjacent even placed bit.
Constrain: While solving the problem, we have to keep In mind that we cannot use a loop for this problem, we have to execute our code in O(1) time complexity only.
Example
Input − n = 10011110
输出 - 在交换偶数位置位和奇数位置位之后，
the binary number obtained is: 01101101
Input − n = 10011110
输出 - 在交换偶数位置位和奇数位置位之后，
the binary number obtained is: 01101101
Explanation −
让我们考虑前面的例子以更好地理解。
n = 10011110 Even position bits in n are E – 1 x 0 x 1 x 1 x Odd position bits in n are O – x 0 x 1 x 1 x 0
For the result, we want the even position bits at the odd position and vice-versa
For even position bits at odd position,
We need to right shift the even position by one position.
因此，对于偶数位置的位，我们只需将 E >> 1 来获取所需的位置。
Similarly, we have to left shift the odd position bits by one position to get the desired position of odd bits.
所以，对于奇数位，我们只需要将O
Now the next problem is to extract the odd and even position bits.
正如我们所知，
0x55 = 01010101 in which every only odd position bits are set ( non 0 ). 0xAA = 10101010 in position bits are set. which, only odd
Hence to extract E from n, we just need to perform
E = n & 0xAA
Similarly, to extract O from n, we need to perform-
O = n & 0x55
Now, to find the swapped output,
步骤
涉及的步骤为-
E >> 1
O
Now, we combine E and O using or operation.
Hence our result will be – Result = ( E >> 1 | O
Example
的中文翻译为：示例
这种方法的代码表示如下：
#includeusing namespace std; unsigned int swapbits(unsigned int n) { unsigned int E = n & 0xAA ; unsigned int O = n & 0x55 ; unsigned int result = (E >> 1)|(O << 1); return result; } int main() { unsigned int n = 14; cout << "After swapping the even position bits with off position bits, the binary number obtained is " << swapbits(n) << endl; return 0; // code is contributed by Vaishnavi tripathi }
Output
After swapping the even position bits with off position bits, the binary number obtained is 13
时间复杂度 - 这种方法的时间复杂度为O(1)。
空间复杂度 - 我们没有使用任何额外的空间。辅助空间复杂度为O(1)。
