标签：dbi put src rac mon nts pow ica fir
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:
Approach #1: DFS + backtracking. [C++]
class Solution {
public:
vector<string> readBinaryWatch(int num) {
init();
vector<string> ans;
for (int i = 0; i <= num; ++i) {
set<int> h;
set<int> m;
hours(i, h, 0);
minutes(num-i, m, 0);
string temp;
for (int i : h) {
for (int j : m) {
//cout << i << ‘ ‘ << j << endl;
if (i >= 12 || j >= 60) continue;
temp = to_string(i);
temp += ":";
if (j < 10) temp += "0" + to_string(j);
else temp += to_string(j);
ans.push_back(temp);
}
}
}
//sort(ans.begin(), ans.end());
return ans;
}
private:
vector<pair<int, bool>> hour;
vector<pair<int, bool>> minute;
void init() {
for (int i = 0; i < 4; ++i)
hour.push_back({pow(2, i), true});
for (int i = 0; i < 6; ++i)
minute.push_back({pow(2, i), true});
}
void hours(int n, set<int>& h, int h_) {
if (n == 0) {
h.insert(h_);
return;
}
for (auto &i : hour) {
if (i.second) {
i.second = false;
hours(n-1, h, h_+i.first);
i.second = true;
}
}
}
void minutes(int n, set<int>& m, int m_) {
//cout << "m_ = " << m_ << " ";
if (n == 0) {
m.insert(m_);
return;
}
for (auto &j : minute) {
if (j.second) {
j.second = false;
minutes(n-1, m, m_+j.first);
j.second = true;
}
}
}
};
标签：dbi put src rac mon nts pow ica fir
原文地址：https://www.cnblogs.com/ruruozhenhao/p/10353376.html
