LeetCode 190 :Reverse Bits

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

Credits:

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

分析：

题目要求将一个整数的二进制翻转，如0x00000001 翻转成0x80000000。这个与10进的翻转类似。

代码如下：

uint32_t reverseBits(uint32_t n) {
int result = 0;
for(int i=0;i<31; i++){
//取最后一位，然后右移
int tmp = n & 0x01;
n = n >> 1;
//先左移一位，与上最后一位
result = (result<<1) | tmp;
}
return result;
}