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 as00111001011110000010100101000000).

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

思路一：

这题目一看就知道肯定要用位运算，但是要考虑到如果最高位是1的话，再右移位运算的时候左边会补1，所以要单独处理一下。

public int reverseBits(int n) {
        int ret = 0;
        int factor = 1;
        factor = factor << 30;
        int tmp = 2;
        //处理最高位
        if ((factor<<1 & n) != 0) {
            ret += 1;
        }
        for(int i=0; i<31; i++) {
            if ((factor & n) != 0) {
                ret += tmp;
            }
            tmp = tmp << 1;
            factor = factor >> 1;
        }
        return ret;
    }

思路二：

把位运算运用到极致的方法。用(n >> i) & 1实现取位，用或运算实现加和。

public int reverseBits2(int n) {
        int ret = 0;
        for(int i=0; i<32; i++) {
            int bit = (n >> i) & 1;
            ret |= bit << (31 - i);
        }
        return ret;
    }