Leetcode Binary Watch

题意：手表用二进制显示时间，给出亮灯的个数，求可能的时间。

思路：枚举子集的个数。这里数量较小，直接用二进制枚举。

class Solution {public:    vector<string> readBinaryWatch(int num) {        vector<int> h(4, 1);        for(int i = 1; i < h.size(); i ++ ) {            h[i] = h[i - 1] * 2;        }        vector<int> m(6, 1);        for(int i = 1; i < m.size(); ++ i) {            m[i] = m[i - 1] * 2;        }                vector<string> re;        vector<int> myh, mym;        for(int i = 0; i <= num; ++ i) {            myh = dfsH(h, i);            mym = dfsM(m, num - i);            for(int j = 0; j < myh.size(); j ++) {                string hour = std::to_string(myh[j]);                for(int k = 0; k < mym.size(); k ++) {                    string minutes;                    if(mym[k] < 10) minutes = "0" + std::to_string(mym[k]);                    else minutes = std::to_string(mym[k]);                    re.push_back(hour + ":" + minutes);                }            }        }                return re;    }        vector<int> dfsH(vector<int> h, int nums) {        vector<int> re;                //int total = 7;        for(int i = 0; i <= 15; ++ i) {            if(ones(i) == nums && sum(h, i) < 12) {                re.push_back(sum(h, i));            }        }                return re;            }    vector<int> dfsM(vector<int> h, int nums) {        vector<int> re;                //int total = 127;        for(int i = 0; i <= 127; ++ i) {            if(ones(i) == nums && sum(h, i) < 60) { //cout << i << " " << ones(i) << " " << sum(h, i) << endl;                re.push_back(sum(h, i));            }        }                return re;            }        int ones(int n) {        int cout = 0;        while(n) {            cout += n%2;            n /= 2;        }        return cout;    }        int sum(vector<int> n, int ind) {        int re = 0;        int i = 0;        while(ind) {            if(i >= n.size()) return 1000;            if(ind % 2) {                re += n[i];            }            ind /= 2;            i ++; //if(i >= n.size()) return 1000;        }                return re;    }};