版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/qq_38283262/article/details/84226747
Description
Reverse bits of a given 32 bits unsigned integer.
Example
Input: 43261596
Output: 964176192
Explanation: 43261596 represented in binary as 00000010100101000001111010011100,
return 964176192 represented in binary as 00111001011110000010100101000000.
Solution
我们把要翻转的数n从低位到高位依次放入res中，放入的方式就是判断此时n的最低位是0还是1，如果是1就将res++，否则直接无视就好了。放入后我们把n右移一位，使高一位成为最低位以做下一次的判断，而每次放入前都将res左移，使低位的数随着循环都移到了高位。通过这种将原本的低位移动到高位的方式，实现了题目所要求的翻转。
public class Solution {
public int reverseBits(int n) {
if (n == 0) return 0;
int res = 0;
for (int i = 0; i < 32; i++) {
res = res << 1;
if ((n & 1) == 1) res++;
n = n >> 1;
}
return res;
}
}