反转二制

原题

　　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?

题目大意

　　反转一个32位无符号的整数。

解题思路

　　设这个数为k，用一个初值为0的数r保存反转后的结果，用1对k进行求与，其结果与r进行相加，再对k向右进行一位移位，对r向左进行一位移位。值到k的最后一位处理完。

代码实现

算法实现类

public class Solution {

    public int reverseBits(int n) {

        int result = 0;
        for (int i = 0; i < 32; i++) {
            result += n & 1;
            n >>>= 1;
            if (i < 31) {
                result <<= 1;
            }
        }
        return result;
    }
}