190. Reverse Bits

题目：

Reverse bits of a given 32 bits unsigned integer.

For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as00111001011110000010100101000000).

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

Related problem: Reverse Integer

链接： http://leetcode.com/problems/reverse-bits/

题解：

用了比较naive的方法， 最高位shift然后和1与，然后再加到res里。由于是unsigned int，所以 2147483648 (10000000000000000000000000000000) 应该返回 1 (00000000000000000000000000000001)。 还有一些神牛的解法在reference里，大都只要几个instruction就可以完成...比较好的有Bit Twiddling Hacks以及Hacker's Delight，要好好学习。

Time Complexity - O(1)， Space Complexity - O(1)。

public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits(int n) {
        int res = 0;
        
        for(int i = 0; i < 32; i++) {
            if(((n >> i) & 1) == 1) {
                res += (1 << (31 - i));
            }
        }
        
        return res;
    }
}

二刷:

还是naive方法...每个比特位先shift然后再&1， 假如结果等于1，那么我们把1左移相同的位数，加到res里。

Java:

Time Complexity - O(1)， Space Complexity - O(1)。

public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits(int n) {
        int res = 0;
        for (int i = 0; i < 32; i++) {
            if (((n >> i) & 1) == 1) {
                res += (1 << 31 - i);
            }
        }
        return res;
    }
}

三刷:

Bit Manipulation，以后一定要Dive Deep

Java:

public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits(int n) {
        int res = 0;
        for (int i = 31; i >= 0; i--) {
            if (((n >> i) & 1) == 1) {
                res += (1 << (31 - i));
            }
        }
        return res;
    }
}

Reference:

https://graphics.stanford.edu/~seander/bithacks.html

https://graphics.stanford.edu/~seander/bithacks.html#BitReverseObvious

http://www.hackersdelight.org/

http://www.hackersdelight.org/hdcodetxt/reverse.c.txt

https://leetcode.com/discuss/27405/o-1-bit-operation-c-solution-8ms