- Animal Shelter
- Queue via Stacks
- Three in One
Animal Shelter
這道題目就是使用deque再合适不過了。
不過deque的實作可以使用list來實作,這裡實作的方法就是開頭一個指針指向開頭、一個指向結尾。
用開頭的指針來做被收養的操作,如果不是符合要求的話,那麼就建立一個list存儲需要的内容。然後需要往收容所增加動物的時候,就把動物加在連結清單的後面。
具體代碼實作:
class CatDogAsylum {
public:
vector<int> asylum(vector<vector<int> > ope) {
// write code here
deque<int> que;
vector<int> seq;
int ope_size = ope.size();
for(int i = ; i < ope_size; i++) {
// put animal into shelter
if(ope[i][] == ) {
que.push_back(ope[i][]);
}
// adopt anmial
else {
vector<int> pre;
if(ope[i][] == ) {
seq.push_back(que.front());
que.pop_front();
}
else {
// adopt dog
if(ope[i][] == ) {
while(!que.empty()) {
int tp = que.front();
if(tp > ) { seq.push_back(tp); que.pop_front(); break; }
else { que.pop_front(); pre.push_back(tp); }
}
if(!pre.empty()) que.insert(que.begin(), pre.begin(), pre.end());
}
// adopt cat
else {
while(!que.empty()) {
int tp = que.front();
if(tp < ) { seq.push_back(tp); que.pop_front(); break; }
else { que.pop_front(); pre.push_back(tp); }
}
if(!pre.empty()) que.insert(que.begin(), pre.begin(), pre.end());
}
}
}
}
return seq;
}
};
在cc189上的代碼如下:
abstract class Animal {
private int order;
protected String name;
public Animal(String n) {name = n;}
public int getOrder() {return order;}
public boolean isOrderThan(Animal a) {
return this.order < a.getOrder();
}
};
class AnimalQueue{
LinkedList<Dog> dogs = new LinkedList<Dog>();
LinkedList<Cat> cats = new LinkedList<Cats>();
private int oreder = ;
public void enqueue(Animal a) {
a.setOrder(order);
order++;
}
public Animal dequeueAny() {
if(dogs.size() == ) {
return dequeueCats();
}
if(cats.size() == ) {
return dequeueDogs();
}
Dog dog = dog.peek();
Cat cat = cats.peek();
if(dog.isOrderThan(cat)) return deuqueDogs();
else return dequeueCats();
}
public Dog dequeueDogs() {
return dogs.poll();
}
public Cat dequeueCats() {
return cats.poll();
}
}
public class Dog extend Animal {
public Dog(String n) {super(n);}
}
public class Cat extend Animal {
public Cat(String n) {super(n);}
}
Queue via Stacks
這裡做的就是把兩個棧實作為隊列,我的方法比較簡單,就是使用一個棧來存資料,另外一個棧用來做中轉。那麼就可以寫成:
class Solution
{
public:
void push(int node) {
if(!stack1.empty()) {
int tmp;
while(!stack1.empty()) {
tmp = stack1.top();
stack1.pop();
stack2.push(tmp);
}
stack1.push(node);
while(!stack2.empty()) {
tmp = stack2.top();
stack2.pop();
stack1.push(tmp);
}
}
else stack1.push(node);
}
int pop() {
if(!stack1.empty()) {int res = stack1.top(); stack1.pop(); return res;}
else return ;
}
private:
stack<int> stack1;
stack<int> stack2;
};