【LeetCode】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 as 00111001011110000010100101000000).

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

提示：

这道题比较简单，考的就是按位操作，可以借助于STL的bitset，也可以直接通过位运算完成题目的要求。

代码：

使用bitset:

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        bitset<32> bits(n);
        int i = 0, j = 31, tmp;
        for (; i < j; ++i, --j) {
            tmp = bits[i];
            bits[i] = bits[j];
            bits[j] = tmp;
        }
        unsigned long l = bits.to_ulong();
        return (uint32_t)l;
    }
};

直接按位操作：

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        uint32_t m = 0;
        for (int i = 0; i < 32 ; ++i, n = n >> 1)
            m = (m << 1) + (n & 1);
        return m;
    }
};