problem

​​190. Reverse Bits​​

 solution1:

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        uint32_t res = 0;
        for(int i=0; i<32; i++)
        {
            res = res*2 + n%2;
            n /=2;
        }
        return res;
    }
};

solution2

使用二进制的移位特性实现。

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        n = (n >> 16) | (n << 16);
        n = ((n & 0xff00ff00) >> 8) | ((n & 0x00ff00ff) << 8);
        n = ((n & 0xf0f0f0f0) >> 4) | ((n & 0x0f0f0f0f) << 4);
        n = ((n & 0xcccccccc) >> 2) | ((n & 0x33333333) << 2);
        n = ((n & 0xaaaaaaaa) >> 1) | ((n & 0x55555555) << 1);
        return n;
    }
};

for 8 bit binary number abcdefgh, the process is as follow:

abcdefgh -> efghabcd -> ghefcdab -> hgfedcba

1. ​​Leetcode_Reverse Bits​​;