LeetCode 401. Binary Watch

题目描述：

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".
思路：有2个思路
1、固定灯数，尝试所有的分配（时针h，分针num-h），用回溯法求解出所有时针和分针的可能能，然后全排列
2、遍历所有时间（0:00~11:59），输出其二进制中‘1’的个数和为num的时间
第一种方法比较复杂，需要用回溯算法，回溯的目标是实现list中任意k个数的和。（因为本题值比较少，也可以使用穷举）
第二种方法比较简单，可以用python oneline实现，
Solution1：

class Solution {
public:
string conver2str(int x) {
stringstream ss;
ss << x;
string s = ss.str();
return s;
}

vector<string> readBinaryWatch(int num) {
vector<int> hours = { 1,2,4,8 };
vector<int> minutes = { 1,2,4,8,16,32 };
vector<string> res;
for (int h = 0; h <= 4 && h <= num; h++) {
vector<int> possibleH = possibleValus(hours, h);
vector<int> possibleM = possibleValus(minutes, num - h);
for (int i=0;i<possibleH.size();i++){
if(possibleH[i]>11)
continue;
for (int j = 0; j < possibleM.size(); j++) {
if(possibleM[j]<60){
string sh = conver2str(possibleH[i]);
string sm = conver2str(possibleM[j]);
if (possibleM[j] < 10)
sm = "0" + sm;
res.push_back(sh + ":" + sm);
}
}
}
}
return res;
}
void getSum(vector<int> l, int count, int pos, int sum, vector<int> &res) {
if (count == 0) {
res.push_back(sum);
return;
}
for (int i = pos; i < l.size(); i++) {
getSum(l, count - 1, i + 1, sum + l[i], res);
}
}

vector<int> possibleValus(vector<int> l, int n) {
vector<int> res;
getSum(l, n, 0, 0, res);
return res;
}
};

Solution2（python）：

return ["%d:%02d"%(h,m) for h in range(12) for m in range(60) if (bin(h)+bin(m)).count('1') == x]