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

思路：
将32位的无符号整数n右移32次，每次移动取出最低位，赋值给另一个变量result, 对应于n的右移，result对应地左移32次。（result需要先右移，再将n的最低位赋值给result，否则当输入n=2147483648时,会溢出，因为java的int类型是有符号的，能表示的最大的数是2147483647）。

public class Solution {    // you need treat n as an unsigned value    public int reverseBits(int n) {        int result = 0;        for (int i=0; i<32; i++){            // 取出n的最低位            int least = n & 0x01;            // n右移一位            n = n >> 1;            // 下面两个步骤需要注意：一定要先左移，再赋值。            // 否则当n=2147483648时，result = result | least 会溢出。            // 将result左移一位            result = result << 1;            // 将n的最低位加到result最后一位            result = result | least;        }        return result;    }}