leetcode 190. 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

这道题很简答，就是做一个移位操作，其实也可以把整数转化二进制的字符串，然后反转字符串。

像这样的题很可能面试遇到。

代码如下：

public class Solution{    /*     * 由于Java中没有无符号数的概念，所以要注意一下，下面是Java的移位运算符     * << 左移运算符，num << 1,相当于num乘以2     * >> 右移运算符，num >> 1,相当于num除以2     * >>> 无符号右移，忽略符号位，空位都以0补齐     *      * */    // you need treat n as an unsigned value    public int reverseBits(int n)     {        int res=0;        for(int i=0;i<32;i++)        {            if( (n&1)==1 )            {                res = (res<<1) +1;                n = n>>>1;            }else            {                res = res<<1;                n = n>>>1;            }        }        return res;    }}

下面是C++的做法，就是直接作位运算

代码如下：

#include <iostream>#include <vector>#include <string>#include <map>#include <cmath>#include <algorithm>using namespace std;class Solution {public:    uint32_t  reverseBits(uint32_t n)    {        uint32_t result = 0;        for (int i = 0; i < 32; i++)        {            result = (result << 1) + (n >> i & 1);        }        return result;    }};