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".
这道题考察我们二进制表，说实话，博主对二进制表无感，感觉除了装b没啥其他的作用，谁会看个时间还要算半天啊，但是这并不影响我们做题，我们首先来看一种写法很简洁的解法，这种解法利用到了bitset这个类，可以将任意进制数转为二进制，而且又用到了count函数，用来统计1的个数。那么时针从0遍历到11，分针从0遍历到59，然后我们把时针的数组左移6位加上分针的数值，然后统计1的个数，即为亮灯的个数，我们遍历所有的情况，当其等于num的时候，存入结果res中，参见代码如下：
class Solution {
public:
vector<string> readBinaryWatch(int num) {
vector<string> res;
for (int h = ; h < ; ++h) {
for (int m = ; m < ; ++m) {
if (bitset<>((h << ) + m).count() == num) {
res.push_back(to_string(h) + (m < ? ":0" : ":") + to_string(m));
}
}
}
return res;
}
};
上面的方法之所以那么简洁是因为用了bitset这个类，如果我们不用这个类，那么应该怎么做呢？这个灯亮问题的本质其实就是在n个数字中取出k个，那么就跟之前的那道Combinations一样，我们可以借鉴那道题的解法，那么思路是，如果总共要取num个，我们在小时集合里取i个，算出和，然后在分钟集合里去num-i个求和，如果两个都符合题意，那么加入结果中即可，参见代码如下：
解法二：
class Solution {
public:
vector<string> readBinaryWatch(int num) {
vector<string> res;
vector<int> hour{, , , }, minute{, , , , , };
for (int i = ; i <= num; ++i) {
vector<int> hours = generate(hour, i);
vector<int> minutes = generate(minute, num - i);
for (int h : hours) {
if (h > ) continue;
for (int m : minutes) {
if (m > ) continue;
res.push_back(to_string(h) + (m < ? ":0" : ":") + to_string(m));
}
}
}
return res;
}
vector<int> generate(vector<int>& nums, int cnt) {
vector<int> res;
helper(nums, cnt, , , res);
return res;
}
void helper(vector<int>& nums, int cnt, int pos, int out, vector<int>& res) {
if (cnt == ) {
res.push_back(out);
return;
}
for (int i = pos; i < nums.size(); ++i) {
helper(nums, cnt - , i + , out + nums[i], res);
}
}
};
下面这种方法就比较搞笑了，是博主在没法想出上面两种方法的情况下万般无奈使用的，你个二进制表再叼也就72种情况，全给你列出来，然后采用跟上面那种解法相同的思路，时针集合取k个，分针集合取num-k个，然后存入结果中即可，参见代码如下：
解法三：
class Solution {
public:
vector<string> readBinaryWatch(int num) {
vector<vector<int>> hours{{},{,,,},{,,,,},{,}};
vector<vector<int>> minutes{{},{,,,,,},{,,,,,,,,,,,,,,},{,,,,,,,,,,,,,,,,,,,},{,,,,,,,,,,,,,},{,,,}};
vector<string> res;
for (int k = ; k <= num; ++k) {
int t = num - k;
if (k > || t > ) continue;
for (int i = ; i < hours[k].size(); ++i) {
for (int j = ; j < minutes[t].size(); ++j) {
string str = minutes[t][j] < ? "" + to_string(minutes[t][j]) : to_string(minutes[t][j]);
res.push_back(to_string(hours[k][i]) + ":" + str);
}
}
}
return res;
}
};
参考资料：
https://discuss.leetcode.com/topic/59401/straight-forward-6-line-c-solution-no-need-to-explain