401. Binary Watch

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".

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;
            } 
        }
    }
};

  

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转载自www.cnblogs.com/ruruozhenhao/p/10353376.html
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