1 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

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

 

Hide Tags
 Bit Manipulation
 
2 分析
通过题191的思路，我们可以获得每位的数据，这样反转也就很方便了，刚开始以为要存起来，反过来赋值一遍，结果不用，两个移位的方向不一样就行了。
ps java里面向高位移位时，直接丢弃，不保存。
 
3 代码
    public int reverseBits(int n) {
        int reverseN = 0;
        int b;
        for (int i = 0; i < 32; i++) {
            b =  n&1;
            reverseN = (reverseN << 1) + b;
            n = n >> 1;
        }

另外，有个问题，不定义b的话，直接用

 reverseN = (reverseN << 1) + n&1;得不到结果，结果全部是0，原因未知。。