[LeetCode] Binary Watch 二进制表
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".
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 = 0; h < 12; ++h) { for (int m = 0; m < 60; ++m) { if (bitset<10>((h << 6) + m).count() == num) { res.push_back(to_string(h) + (m < 10 ? ":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{8, 4, 2, 1}, minute{32, 16, 8, 4, 2, 1}; for (int i = 0; i <= num; ++i) { vector<int> hours = generate(hour, i); vector<int> minutes = generate(minute, num - i); for (int h : hours) { if (h > 11) continue; for (int m : minutes) { if (m > 59) continue; res.push_back(to_string(h) + (m < 10 ? ":0" : ":") + to_string(m)); } } } return res; } vector<int> generate(vector<int>& nums, int cnt) { vector<int> res; helper(nums, cnt, 0, 0, res); return res; } void helper(vector<int>& nums, int cnt, int pos, int out, vector<int>& res) { if (cnt == 0) { res.push_back(out); return; } for (int i = pos; i < nums.size(); ++i) { helper(nums, cnt - 1, i + 1, out + nums[i], res); } } };