遍历HashMap并获取对象中值的“最低”整数

Question

    for (Entry<String, Data> entry : list.entrySet()) {
        if(entry.getValue().getRoom() == 1){
            if(entry.getValue().getName().equalsIgnoreCase("RED")){
                 entry.getValue().getPosition() // need to get the lowest free number
                 // between the range of 1-6
            } 
        }
    }

How to get the lowest free spot of the getPosition in this situation. getPosition values are between 1-6 and there are only one of each value Room = 1 and Name = RED. For example if 1,3,4,6 exists in getPosition(with room=1 and name=red) then the output should be 2. That is the lowest number that is free in getPosition in the specific combination. Hope u can help me out.

Answer 1

Well, it sounds like the simplest approach would be something like:

boolean[] taken = new boolean[7]; //(0-6 inclusive)
// You were never using the key as far as I could see...
for (Data data : list.values()) {
   if (data.getRoom() == 1 && data.getName().equalsIgnoreCase("RED")) {
       taken[data.getPosition()] = true;
   }
}

for (int i = 1; i <= 6; i++) {
    if (!taken[i]) {
        return i;
    }
}

// Do whatever you want if there are no free positions...