Binary Watch【二进制手表时间转换为普通时间】

PROBLEM:

---

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".

SOLVE:

class Solution {
public:
    vector<string> readBinaryWatch(int num) {
    	//pay attention to 60~63 of minite and 12~15 of hour
    	vector<string> res;
    	int h = 0, m = 0;
    	vector<vector<int>> hourKind = kindOfhour();
    	vector<vector<int>> minKind = kindOfmin();
    	for (int i = 0;i <= 4 && i <= num;++i) {
    		//i is count of hour LED, j is count of min LED, should promise 0<=i<=4 & 0<=j<=6
	    	int j = num - i;
	    	if (j > 6) continue;    //num may be 7,8,9,10, it will cause wrong when i equal to 0
	    	for (int h : hourKind[i])
	    		for (int m : minKind[j])
	    			res.push_back(numToTime(h, m));
    	}
    	return res;
    }
private:
    string numToTime(int h, int m) {
	    //transform (h,m) to string in the form of "X:XX"
    	string res;
    	res = to_string(h) + ":";
    	if (m<10)
    		res = res + '0' + to_string(m);
    	else
    		res += to_string(m);
    	return res;
    }
    vector<vector<int>> kindOfhour(void) {
    	//res[i],i is the num of LED, res[i] consist of all impossible hours
    	vector<vector<int>> res(5, vector<int>());
    	for (int i = 0;i<12;i++) {
    		int Index = 0;
    		for (int tmp = i;tmp>0;tmp = tmp >> 1)
    			if (tmp % 2 == 1)
    				Index++;
    		res[Index].push_back(i);
    	}
    	return res;
    }
    vector<vector<int>> kindOfmin(void) {
    	//res[i],i is the num of LED, res[i] consist of all impossible minitues
    	vector<vector<int>> res(7, vector<int>());
    	for (int i = 0;i<60;i++) {
    		int Index = 0;
    		for (int tmp = i;tmp>0;tmp = tmp >> 1)
    			if (tmp % 2 == 1)
    				Index++;
    		res[Index].push_back(i);
    	}
    	return res;
    }
}；

分析：这里一个技巧是在求n个LED时可能的组合数，反过来对组合数遍历，放入对应LED组合中。