力扣 190.颠倒二进制位

被简单题打败了，自己写的再原n上操作遍历交换 1ms，
题解一，建立一个新的载体，遍历复制 0ms
最可怕的是题解二：位运算分治 ， 位运算果然是最难懂的代码。。

题解一（0ms）

public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits(int n) {
        int rev = 0;
        for (int i = 0; i < 32 && n != 0; ++i) {
            rev |= (n & 1) << (31 - i);
            n >>>= 1;
        }
        return rev;
    }
}

题解二（0ms）

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;
    }
}