Implement Stack using Queues
描述
Implement the following operations of a stack using queues.
- push(x) -- Push element x onto stack.
- pop() -- Removes the element on top of the stack.
- top() -- Get the top element.
- empty() -- Return whether the stack is empty.
Notes:
- You must use only standard operations of a queue -- which means only push to back, peek/pop from front, size, and is empty operations are valid.
- Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as you use only standard operations of a queue.
- You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).
分析
可以用两个队列，q和tmp，q存放元素，tmp用来作中转。
push(x)，先将xpush 到tmp，然后把q中的元素全部弹出来，存入tmp，最后切换q和tmppop()，直接将q的队首元素弹出来即可
该算法push的算法复杂度是O(n), pop的算法复杂度O(1)。
另个一个方法是，让pop是O(n), push是O(1)，思路很类似，就不赘述了。
代码
// Implement Stack using Queues
class MyStack {
// Push element x onto stack.
// Time Complexity O(n)
public void push(int x) {
tmp.offer(x);
while (!q.isEmpty()) {
final int e = q.poll();
tmp.offer(e);
}
// swap q and tmp
Queue<Integer> temp = tmp;
tmp = q;
q = temp;
}
// Removes the element on top of the stack.
// Time Complexity O(1)
public void pop() {
q.poll();
}
// Get the top element.
public int top() {
return q.peek();
}
// Return whether the stack is empty.
public boolean empty() {
return q.isEmpty();
}
private Queue<Integer> q = new LinkedList<>();
private Queue<Integer> tmp = new LinkedList<>();
}