[LeetCode] Reverse Bits 翻转位

Reverse bits of a given 32 bits unsigned integer.

For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as00111001011110000010100101000000).

Follow up:
If this function is called many times, how would you optimize it?

Related problem: Reverse Integer

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

这道题又是在考察位操作Bit Operation，LeetCode中有关位操作的题也有不少，比如 Repeated DNA Sequences 求重复的DNA序列， Single Number 单独的数字,   Single Number II 单独的数字之二 ，和 Grey Code 格雷码 等等。跟上面那些题比起来，这道题简直不能再简单了。那么对于这道题，我们只需要把要翻转的数从右向左一位位的取出来，然后加到新生成的数的最低位即可，代码如下：

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        uint32_t res = 0;
        for (int i = 0; i < 32; ++i) {
            if (n & 1 == 1) {
                res = (res << 1) + 1;
            } else {
                res = res << 1;
            }
            n = n >> 1;
        }
        return res;
    }
};

当然，也可以用下面这种写法让代码更简洁一些：

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        uint32_t res = 0;
        for (int i = 0; i < 32; ++i) {
            res |= ((n >> i) & 1) << (31 - i);
        }
        return res;
    }
};

本文转自博客园Grandyang的博客，原文链接：翻转位[LeetCode] Reverse Bits ，如需转载请自行联系原博主。