[LeetCode] 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.

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"]

這是一道很簡單的題，基本上就是十進制和二進制的轉換而已。但是讓我意識到自己還有很多知識的盲區。

這是我寫的Accepted代碼，真的是搓到不行：

class Solution {
public:
    vector<pair<int,int>> hour;
    vector<pair<int,int>> minute;
    void initAll()
    {
        for(int i=0;i<12;i++)
        {
            int cur=i,temp=0;
            while(cur>0)
            {
                if(cur%2==1)temp++;
                cur/=2;
            }
            hour.push_back(make_pair(i,temp));
        }
        for(int i=0;i<60;i++)
        {
            int cur=i,temp=0;
            while(cur>0)
            {
                if(cur%2==1)temp++;
                cur/=2;
            }
            minute.push_back(make_pair(i,temp));
        }
    }
    vector<int> getHour(int num)
    {
        vector<int> temp;
        for(vector<pair<int,int>>::iterator iter=hour.begin();iter!=hour.end();iter++)
        {
            pair<int,int> tempPair=*iter;
            if(tempPair.second==num)temp.push_back(tempPair.first);
        }
        return temp;
    }
    vector<int> getMinute(int num)
    {
        vector<int> temp;
        for(vector<pair<int,int>>::iterator iter=minute.begin();iter!=minute.end();iter++)
        {
            pair<int,int> tempPair=*iter;
            if(tempPair.second==num)temp.push_back(tempPair.first);
        }
        return temp;
    }
    vector<string> readBinaryWatch(int num) {
        initAll();
        vector<string> ans;
        for(int i=0;i<=num;i++)
        {
            int j=num-i;
            if(i>4||j>6)continue;
            vector<int> tempHour=getHour(i);
            vector<int> tempMinute=getMinute(j);
            int len1=tempHour.size();
            int len2=tempMinute.size();
            for(int p=0;p<len1;p++)
            {
                for(int q=0;q<len2;q++)
                {
                    string tempans=to_string(tempHour[p]);
                    tempans+=":";
                    if(tempMinute[q]<=9)tempans+="0";
                    tempans+=to_string(tempMinute[q]);
                    ans.push_back(tempans);
                }
            }
        }
        return ans;
    }
};

然後這是看到的別的大佬寫的：

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

你們感受一下這兩份代碼的差距......

我只能說，bitset真乃神器也。路漫漫其修遠兮，吾將上下而求索。