[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 > ) return res;
     for (int i = ; i < ( << ); i++) {
         if(__builtin_popcount(i) == k) {
             int h = i >> ;
             if(h > ) continue;
             int m = i & (( << ) - );
             if(m > ) continue;
             stringstream ss;
             ss << h; ss << ":";
             if(m < ) ss << "";
             ss << m;
             string x = ss.str();
             //cout << x << endl;
             res.push_back(x);
         }
     }
     return res;
 }
    vector<string> readBinaryWatch(int num) {
         return work(num);
     }
 };