题目:
Reverse bits of a given 32 bits unsigned integer.
For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as 00111001011110000010100101000000).
Follow up:
If this function is called many times, how would you optimize it?
Related problem: Reverse Integer
思路:
也是一道easy级别的题目,算法方面没有什么难度,但我这里提供一个非常精简的代码片段:申请一个变量ret,然后每次让其左移1位,和n右移1位的结果进行或操作,这样每次就能得到n的最低位。四行代码就搞定了。
代码:
class Solution {
public:
uint32_t reverseBits(uint32_t n) {
uint32_t ret = 0;
for(int i = 0; i < 32; ++i)
ret = (ret << 1) | ( (n >> i) & 1); // update rightmost bit
return ret;
}
};