原题传送门

1. 题目描述

2. Solution 1

1、思路分析
逐位颠倒
We first initialize result to 0. We then iterate from
0 to 31 (an integer has 32 bits). In each iteration:
We first shift result to the left by 1 bit.
Then, if the last digit of input n is 1, we add 1 to result. To
find the last digit of n, we just do: (n & 1)
Example, if n=5 (101), n&1 = 101 & 001 = 001 = 1;
however, if n = 2 (10), n&1 = 10 & 01 = 00 = 0).

Finally, we update n by shifting it to the right by 1 (n >>= 1). This is because the last digit is already taken
care of, so we need to drop it by shifting n to the right by 1.

2、代码实现

package Q0199.Q0190ReverseBits;

/*
    We first initialize result to 0. We then iterate from
    0 to 31 (an integer has 32 bits). In each iteration:
    We first shift result to the left by 1 bit.
    Then, if the last digit of input n is 1, we add 1 to result. To
    find the last digit of n, we just do: (n & 1)
    Example, if n=5 (101), n&1 = 101 & 001 = 001 = 1;
    however, if n = 2 (10), n&1 = 10 & 01 = 00 = 0).

    Finally, we update n by shifting it to the right by 1 (n >>= 1). This is because the last digit is already taken
    care of, so we need to drop it by shifting n to the right by 1.
 */
public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits(int n) {
        if (n == 0) return 0;
        int result = 0;
        for (int i = 0; i < 32; i++) {
            result <<= 1;   // 把当前结果保存到result的最低位
            if ((n & 1) == 1) result++;
            n >>= 1;
        }
        return result;
    }
}

3、复杂度分析
时间复杂度: O(32) = O(1)
空间复杂度: O(1)

3. Solution 2

1、思路分析
位运算分治
若要翻转一个二进制串，可以将其均分成左右两部分，对每部分递归执行翻转操作，然后将左半部分拼在右半部分的后面，即完成了翻转。
由于左右两部分的计算方式是相似的，利用位掩码和位移运算，可以自底向上地完成这一分治流程。

对于递归的最底层，需要交换所有奇偶位: 取出所有奇数位和偶数位；将奇数位移到偶数位上，偶数位移到奇数位上。类似地，对于倒数第二层，每两位分一组，按组号取出所有奇数组和偶数组，然后将奇数组移到偶数组上，偶数组移到奇数组上。

2、代码实现

public class Solution {
    private static final int M1 = 0x55555555; // 01010101010101010101010101010101
    private static final int M2 = 0x33333333; // 00110011001100110011001100110011
    private static final int M4 = 0x0f0f0f0f; // 00001111000011110000111100001111
    private static final int M8 = 0x00ff00ff; // 00000000111111110000000011111111

    public int reverseBits(int n) {
        n = n >>> 1 & M1 | (n & M1) << 1;
        n = n >>> 2 & M2 | (n & M2) << 2;
        n = n >>> 4 & M4 | (n & M4) << 4;
        n = n >>> 8 & M8 | (n & M8) << 8;
        return n >>> 16 | n << 16;
    }
}

3、复杂度分析
时间复杂度: O(1)
空间复杂度: O(1)