[日常刷题]leetcode D31_凯勒数学答案-优快云博客

本文深入探讨了二进制手表时间表示的算法实现，以及将整数转换为十六进制字符串的方法，包括处理负数的二进制补码表示。文章提供了C++代码示例，展示了如何利用位操作和STL bitset进行高效计算。

文章目录

401. Binary Watch

A binary watch has 4 LEDs on the top which represent the hours (0-11), and the 6 LEDs on the bottom represent the minutes (0-59).

Each LED represents a zero or one, with the least significant bit on the right.
在这里插入图片描述
For example, the above binary watch reads “3:25”.

Given a non-negative integer n which represents the number of LEDs that are currently on, return all possible times the watch could represent.
Example:

Input: n = 1
Return: ["1:00", "2:00", "4:00", "8:00", "0:01", "0:02", "0:04", "0:08", "0:16", "0:32"]

Note:

The order of output does not matter.
The hour must not contain a leading zero, for example “01:00” is not valid, it should be “1:00”.
The minute must be consist of two digits and may contain a leading zero, for example “10:2” is not valid, it should be “10:02”.

Solution in C++：

关键点：

位数与数字的对应

思路：

我的思路有点复杂，其实总的来说是不知道STL的使用的锅。先说下我的大概思路，首先构造分钟和时钟分别不同位数对应的不同字符串。然后根据时钟最多4位，分钟最多6位的限制，进行整数划分。其中包含了几个操作，首先是初始化了tables用于存储每个数字含有几个bit的1；bits函数用于计算num中含有bit的数目，然后根据整数划分直接读取map中的值就OK了。
看到discuss里面有直接关于bit的STL bitset我真的是学到了，这题花了我一个多小时。。。今天注定刷不了几题了。

方法一：手动求num->bits

vector<int> tables;
    
    int bits(int num){
        int result = 0;
        while(num){
            ++result;
            num &= (num - 1);
        }
        return result;
    }
    
    void init(vector<int> &t){
        for(int i = 0; i <= 59; ++i)
            tables.push_back(bits(i));
    }
    
    vector<string> readBinaryWatch(int num) {
                
        init(tables);
        vector<string> result, str;
        map<int,vector<string>> hours, minutes;
        
        // 计算时钟字符串
        for(int i = 0; i <= 4; ++i){
            str.clear();
            vector<int> tmp;
            for(int j = 0; j <= 11; ++j)
                if(tables[j] == i)
                    tmp.push_back(j);
            for(auto t : tmp){
                str.push_back(to_string(t) + ":");
            }
            hours[i] = str;
        }
        
        // 计算分钟字符串
        for(int j = 0; j <= 6; ++j){
            str.clear();
            vector<int> tmp;
            for(int i = 0; i <= 59; ++i)
                if(tables[i] == j)
                    tmp.push_back(i);
            for(auto t : tmp){
                if (t < 10)
                    str.push_back("0"+to_string(t));
                else    
                    str.push_back(to_string(t));
            }
            minutes[j] = str;
        }
         
        // 整数划分 时钟部分：0-4  分钟：0-6
        for(int i = 0; i <= 4; ++i){
            int j = num - i;
            if ((j <= 6) && (j >= 0)){
                for(auto h : hours[i]){
                    for(auto m : minutes[j])
                        result.push_back(h + m);
                }      
            }
        }
        
        return result;
    }

方法二：bitset

vector<string> readBinaryWatch(int num) {
        vector<string> result;
        for(int h = 0; h < 12; ++h){
            for (int m = 0; m < 60; ++m){
                if (bitset<10>(h<<6 | m).count() == num)
                    result.emplace_back(to_string(h) + (m < 10?":0":":") + to_string(m));
            }
        }
        
        return result;
    }

405. Convert a Number to Hexadecimal

Given an integer, write an algorithm to convert it to hexadecimal. For negative integer, two’s complementmethod is used.

Note:

All letters in hexadecimal (a-f) must be in lowercase.
The hexadecimal string must not contain extra leading 0s. If the number is zero, it is represented by a single zero character '0'; otherwise, the first character in the hexadecimal string will not be the zero character.The hexadecimal string must not contain extra leading 0s. If the number is zero, it is represented by a single zero character ‘0’; otherwise, the first character in the hexadecimal string will not be the zero character.
The given number is guaranteed to fit within the range of a 32-bit signed integer.The given number is guaranteed to fit within the range of a 32-bit signed integer.
You must not use any method provided by the library which converts/formats the number to hex directly.You must not use any method provided by the library which converts/formats the number to hex directly.

Example 1:

Input:
26

Output:
"1a"

Example 2:

Input:
-1

Output:
"ffffffff"

Solution in C++：

关键点：

计算机中数的表示

思路：

看了2的补码wiki，得到计算负数 $n u m$ 的补码的数值大小为 $2^n+num$ ，但是int的范围是[- $2^{31}$ , $2^{31}-1$ ]，所以这里表示的时候需要通过long long类型的变量来转存。这里的思路就是小于0时令变量为 $2^n+num$ 否则为 $n u m$ 即可。其他按进制转换操作来。
看到discuss里面很多人都是通过强制类型转换为unsigned int来实现的。

方法一： $2^n+num$

string toHex(int num) {
        
        if (num == 0)
            return "0";
        
        string result; 
        long long tmp = num;
        
        // 负数
        if (num < 0){
             tmp = pow(2,32) + num;
        }
           
        while(tmp){
            int n = tmp % 16;
            if (n >= 10){
                char ch = 'a' + n - 10;
                result = ch + result;
            } else{
                result = to_string(n) + result;
            }
            tmp /= 16;
        }    
            
        return result;
    }

方法二：强制类型转换为unsigned

string helper(unsigned int num){
        string result; 
   
        while(num){
            int n = num % 16;
            if (n >= 10){
                char ch = 'a' + n - 10;
                result = ch + result;
            } else{
                result = to_string(n) + result;
            }
            num /= 16;
        }
    
        return result;
    }
    
    string toHex(int num) {
        return helper(num); 
    }