190. Reverse Bits

问题
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.

创建返回值ans。
每次向左移动一位ans，然后取n的尾数进行二进制或运算（相当于在尾部进行不进位的加和）。
然后将n向左移动一位。

• 二进制下的或运算只保留两数间较大的位。(0011 | 0110 = 0111)
• 二进制下的与运算只保留两数间皆为1的位。(0011 & 0110 = 0010)
• 掩码（Mask）是在二进制下进行与运算。以1作为掩码时，前面的31为皆为0，因此进行与运算后只保留最后一位。

因此(n & 1)相当于n的二进制与000…001运算，只保留n的尾数。
然后(ans << 1)向左移动一位，用 | 操作将n的尾数加入ans。

public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits(int n) {
        int ans = 0;
        for(int i = 0; i < 32; i++){
            ans = (ans << 1) | (n & 1);
            n >>= 1;
        }
        return ans;
    }
}