解答的思路：建立一个queue放狗，一个queue放猫。
如下：
import java.util.*;
class Dog{
int time;
int value;
Dog(int a, int b){
time = a;
value = b;
}
}
class Cat{
int time;
int value;
Cat(int a, int b){
time = a;
value = b;
}
} public class CatDogAsylum { public static ArrayList<Integer> asylum(int[][] ope){
ArrayList<Integer> res = new ArrayList();
Queue<Dog> dog = new ArrayDeque();
Queue<Cat> cat = new ArrayDeque();
int time = 0;
for(int i = 0; i < ope.length; i++){ if(ope[i][0] == 1){//push if(ope[i][1] > 0){//dog
Dog tmp = new Dog(time++,ope[i][1]);
dog.add(tmp); }
else if(ope[i][1] < 0){//cat
Cat tmp = new Cat(time++,ope[i][1]);
cat.add(tmp);
}
} else if(ope[i][0] == 2){//pop
if(ope[i][1] == 0){
if(!dog.isEmpty() && ! cat.isEmpty()){
if(dog.peek().time < cat.peek().time){
res.add(dog.peek().value);
dog.poll();
}
else{
res.add(cat.peek().value);
cat.poll();
}
}
else if(!dog.isEmpty()){
res.add(dog.peek().value);
dog.poll();
}
else {
res.add(cat.peek().value);
cat.poll();
}
}
else if(ope[i][1] == 1){
if(!dog.isEmpty()){
res.add(dog.peek().value);
dog.poll();
}
}
else if(ope[i][1] == -1){
if(!cat.isEmpty()){
res.add(cat.peek().value);
cat.poll();
}
} } } return res; }
}