[leetcode] 401. Binary Watch

https://leetcode.com/contest/5/problems/binary-watch/

这个题应该是这次最水的，这个题目就是二进制，然后是10位，最大1024，然后遍历，找二进制里面1的个数为满足要求的，然后判断是否超限，题目说的很清楚，小时0到11，分钟0到59，然后转化成要求的格式就行了。

class Solution {
public:
vector<string> work(int k) {
    vector<string> res;
    if(k > 10) return res;
    for (int i = 0; i < (1 << 10); i++) {
        if(__builtin_popcount(i) == k) {
            int h = i >> 6;
            if(h > 11) continue;
            int m = i & ((1 << 6) - 1);
            if(m > 59) continue;
            stringstream ss;
            ss << h; ss << ":";
            if(m < 10) ss << "0";
            ss << m;
            string x = ss.str();
            //cout << x << endl;
            res.push_back(x);
        }
    }
    return res;
}
   vector<string> readBinaryWatch(int num) {
        return work(num);
    }
};