lintcode--二进制时间
class Solution {
public:
/*
* @param : the number of "1"s on a given timetable
* @return: all possible time
*/
vector<string> binaryTime(int num) {
// Write your code here
vector<int>watch;
for(int i=0;i<10;i++){
watch.push_back(0);
}
vector<string>res;
findtime(watch,res,0,num,0);
return res;
}
void findtime(vector<int>&watch,vector<string>&res,int index,int num,int count)
{
if(count==num){//找到满足条件的时间，将其加入数组中
if(array_to_time(watch)!=""){
res.push_back(array_to_time(watch));
}
return;
}
for(int i=index;i<10;i++){
watch[i]=1;
findtime(watch,res,i+1,num,count+1);
watch[i]=0;
}
}
string array_to_time(vector<int>res)
{
int min=0;
int hour=0;
for(int i=0;i<6;i++){
min=min+pow(2,i)*res[i];
}
for(int i=6;i<10;i++){
hour=hour+pow(2,i-6)*res[i];
}
if(hour>=12||min>=60){//由于时间是在0:00--11:59之间，超过这个的就排除掉
string s="";
return s;
}
if(min<10){//由于分钟必须由两位数组成, 可能包含前导零，故min小于10时，我们需要在其前面加0
string s="";
s=s+to_string(hour)+":0"+to_string(min);
return s;
}else{
string s="";
s=s+to_string(hour)+":"+to_string(min);
return s;
}
}
void printvector(vector<int>res){
for(int i=0;i<();i++){
cout<<res[i]<<" ";
}
cout<<endl;
}
};