Leetcode: Reverse Bits

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?

Related problem: Reverse Integer

Reverse Integer那道题会考虑溢出，因为那是reverse each digit，这里不会溢出，因为reverse的是each bit

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++) {
            int bit = (n>>i) & 1;
            res |= bit<<(31-i); 
        }
        return res;
    }
}

 Q：如果该方法被大量调用，或者用于处理超大数据（Bulk data）时有什么优化方法？
A：这其实才是这道题的精髓，考察的大规模数据时算法最基本的优化方法。其实道理很简单，反复要用到的东西记下来就行了，所以我们用Map记录之前反转过的数字和结果。更好的优化方法是将其按照Byte分成4段存储，节省空间。参见这个帖子。How to optimize if this function is called multiple times? We can divide an int into 4 bytes, and reverse each byte then combine into an int. For each byte, we can use cache to improve performance.

// cache
private final Map<Byte, Integer> cache = new HashMap<Byte, Integer>();
public int reverseBits(int n) {
    byte[] bytes = new byte[4];
    for (int i = 0; i < 4; i++) // convert int into 4 bytes
        bytes[i] = (byte)((n >>> 8*i) & 0xFF);
    int result = 0;
    for (int i = 0; i < 4; i++) {
        result += reverseByte(bytes[i]); // reverse per byte
        if (i < 3)
            result <<= 8;
    }
    return result;
}

private int reverseByte(byte b) {
    Integer value = cache.get(b); // first look up from cache
    if (value != null)
        return value;
    value = 0;
    // reverse by bit
    for (int i = 0; i < 8; i++) {
        value += ((b >>> i) & 1);
        if (i < 7)
            value <<= 1;
    }
    cache.put(b, value);
    return value;
}