leetcode 190. Reverse Bits

不念不忘少年蓝@ 2022-06-08 08:22 154阅读 0赞

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;
        }
    };