Title

[the problem link](https://leetcode.com/problems/binary-watch/description/）

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”.

回溯法，用hour<12&&minute<60判断是否ok，回溯到头的标志是n==0,用一个额外数字表示当前进度

Solution

//cpp:
class Solution {
public:
	vector<int> data = { 1, 2, 4, 8, 1, 2, 4, 8, 16, 32 };
	void f(int n, pair<int,int> time,vector<string> &v,int start) {
		if (n == 0) {
			v.push_back(to_string(time.first) + (time.second < 10 ? ":0" : ":") + to_string(time.second));
		}
		for (int i = start; i < data.size(); i++) {
			if (i < 4) {
				time.first += data[i];
				if (time.first < 12) {
					f(n - 1, time, v, i + 1);
				}
				time.first -= data[i];

			}
			else {
				time.second += data[i];
				if (time.second < 60) {
					f(n - 1, time, v, i + 1);
				}
				time.second -= data[i];
			}
		}
	}
	vector<string> readBinaryWatch(int num) {
		vector<string> v;
		f(num, make_pair(0, 0), v, 0);
		for (auto it : v)
			cout << it << endl;
		return v;
	}
};