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?

题目含义：反转一个32位无符号的整数。

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。

 1     public int reverseBits(int n) {
 2         int res = 0;
 3 //        for (int i = 0; i < 32; ++i) {
 4 //            res |= ((n >> i) & 1) << (31 - i);
 5 //        }
 6 //        return res;
 7         for(int i = 0; i < 32; i++)
 8         {
 9             res = (res << 1) + (n & 1);
10             n = n >> 1;
11         }
12         return res;        
13     }