LeetCode 401. Binary Watch 题解-优快云博客

本文探讨了如何解决二进制手表的问题，该问题要求根据激活的LED数量确定所有可能显示的时间。文章详细介绍了生成所有可能的时间组合的过程，并检查这些组合是否构成有效的时间。

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题目描述：

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.

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".

解题思路：

由于一共有十盏灯，给出一个数n，枚举出所有n盏灯可能的情况。首先将10盏灯编号，然后根据n，生成n组合。代码中generate_Comb即为生成组合的函数，通过这种枚举的方法，判断所代表的时间是否合法（0=<h=<11,0=<m<=59）,将合法答案加入ans中即可。

代码展示：

class Solution {
public:
    void generate_Comb(vector<int>&tmp,int n,int r )
    {
		int i=r;
		while(tmp[i]==n-r+i)
		{
			i--;
		}
		tmp[i]++;
		for(int j=i+1;j<=r;j++)
		{
			tmp[j]=tmp[i]+j-i;
		}
    } 
    string ntoh(int n)
    {
        string ans = "";
        if(n==0) 
        {
            ans="0";
            return ans;
        }
            
        while(n>0)
        {
            ans = char(n%10+'0')+ans;
            n/=10;
        }
        
        return ans;
    }
    string ntom(int n)
    {
        bool flag = false;
        if(n<10)
            flag = true;
        string ans = "";
        if(n==0)
        {
            ans="00";
            return ans;
        }
        while(n>0)
        {
            ans = char(n%10+'0')+ans;
            n/=10;
        }
        if(flag) return "0"+ans;
        return ans;
    }
    void cal_time(vector<string>&res,vector<int>tmp)
    {
	int h=0;
    int m=0;
	for(int i=1;i<tmp.size();i++)
	{
		if(tmp[i]<=4)
		{
			h+=(1<<(tmp[i]-1));
			if(h>11) return; 
		}
		else
		{
			m+=(1<<(tmp[i]-4-1));
			if(m>59) return;
		}
	}
	res.push_back(ntoh(h)+":"+ntom(m));
	return;
    }
    vector<string> readBinaryWatch(int num) {
        vector<string> ans;
        if(num==0)
        {
            ans.push_back("0:00");
            return ans;
        }
        vector<int> tmp;
 	    tmp.push_back(0);
 	    for(int i=1;i<=num;i++)
		    tmp.push_back(i);
		cal_time(ans,tmp);
        int numofloop = 1;
        int dw = 1;
        int up =1;
        for(int i=1;i<=num;i++)
        {
             dw*=i;
        }
        for(int i=1;i<=num;i++)
        {
             up*=(10-i+1);
        }
        numofloop = up/dw;
        while(--numofloop)
        {
            generate_Comb(tmp,10,num );
            cal_time(ans,tmp);
        }
        return ans;
        
    }
};