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

这道题不用我们自己去考虑怎么把int 转化为unsigned int，题目已经帮我们自动转化好了

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。(未深究)