# 190. Reverse Bits
Reverse bits of a given 32 bits unsigned integer.

Note:

> Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer's internal binary representation is the same, whether it is signed or unsigned.
> In Java, the compiler represents the signed integers using 2's complement notation. Therefore, in Example 2 above, the input represents the signed integer -3 and the output represents the signed integer -1073741825.

## Example 1:

Input: n = 00000010100101000001111010011100
Output:    964176192 (00111001011110000010100101000000)
Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192 which its binary representation is 00111001011110000010100101000000.

## Example 2:

Input: n = 11111111111111111111111111111101
Output:   3221225471 (10111111111111111111111111111111)
Explanation: The input binary string 11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is 10111111111111111111111111111111.

## 解題思路
提示：此題是位移運算，關鍵在弄懂 << >> 的位移運算 與 num & 1 的操作，才能搞懂題目!

此題只需要把要翻轉的二進制數從右向左一位位的取出來，如果取出來的是1，將結果(res)左移一位並且加上1，
如果取出來的是0，將結果 res 左移一位，然後將n右移一位即可。

Note =>『num & 1 可以取出當前的二進制數(num) 最右邊那個數』。

## Java Code

func reverseBits(num uint32) uint32 {
    var res uint32 = 0

    //因為只有32位，所以只走訪32次就行了
    for i := 0; i < 32; i++ {

    //透過(num & 1)判斷當前二進制數最右邊那個數
        if (num & 1) == 1 { //若是1
            res = (res << 1) + 1 //則在res向左位移一位，並且+1，(位移本身只會補0，但因為此處要是1，所以才+1)

            //printBits(res) debug用 可以查看每個階段的數值
        } else { //若不是零
            res = res << 1 //單純位移就好，因為位移本身就會補0
            //printBits(res) debug用 可以查看每個階段的數值
        }
        num = num >> 1 //右移 把輸入的input右移，代表會在最左邊補0，最後右邊的數就會消失，變成原本的倒數第二個數
		        //  e.g. 10010 右移後變成 01001
    }

    return res
}

//打印二進制數用的 方便看過程
func printBits(x uint32) {
    var slice []string
    for i := 31; i >= 0; i-- {
        if (x & (1 << i)) != 0 {
            slice = append(slice, "1")
        } else {
            slice = append(slice, "0")
        }
	}
    fmt.Println(strings.Join(slice, ""))
}