190. Reverse Bits

Difficulty: Easy

Related Topics: Bit Manipulation

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 . Therefore, in Example 2 above, the input represents the signed integer -3 and the output represents the signed integer -1073741825.

Follow up:

If this function is called many times, how would you optimize it?

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.

Constraints:

• The input must be a binary string of length 32

Solution

Language: Java

public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits(int n) {
        int ans = 0;

        // 想象一下
        // 两条连通的传送带: A 和 B
        // A 一边往右边输送物品    --------->      
        // B 一边往左边输送物品    <---------
        for (int i = 0; i < 31; i++) {
            ans <<= 1;
            // 取出 n 的第 i 位 bit，把它贴到在 ans 上面
            ans |= n >> i & 1;
        }   

        return ans;
    }
}