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?

JAVA程式碼：
方法一：依次遍歷數字的二進位制位，遇到第k位為1，則結果反向（32-1-k）置1

public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits(int n) {
        int result = 0;
        if((n == 0) || (n == 0xffffffff ))
        {
            return n;
        }
        for(int k = 0; k < 32; k++)
        {
            if((n & 0x01) == 1)
            {
              result |= (1
 
 << (32-k-1));
            }
            n = n>>1;
        }
        return result;
    }
}

法二：第二種方法是參考網上的寫法：

public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits(int n) {
           int Int_Size = Integer.SIZE;
        if((n == 0) || (n == 0xffffffff
 
 ))
        {
            return n;
        }
        for(int i = 0; i <  Int_Size/2; i++)
        {
            int j = Int_Size-1-i;

            int low = (n >> i) & 1;
            int high = (n >> j) & 1;

            int a = 1 << i;
            int b = 1 << j;
            if((low ^ high) == 1)
            {
                n = n ^ (a|b);
            }

        }
        return n;
    }
}