Implement Stack using Queues

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).

Update (2015-06-11):

The class name of the Java function had been updated to MyStack instead of Stack.

Credits:

Special thanks to @jianchao.li.fighter for adding this problem and all test cases.

用队列实现栈，因为用队列实现栈，在pop或者top操作时，难免会出现需要O（n）时间复杂度的情况，开始的时候还打算用两个队列来避免这种情况，但是后来发现，貌似O（n）是不可避免的了。

好多实现都采用了双队列的方式实现，但是在讨论区发现了用单队列实现的方法，通俗易懂，整理后的方法如下：

class MyStack {

private Queue<Integer> queue=new LinkedList<Integer>();

// Push element x onto stack.
public void push(int x) {
queue.add(x);
for(int i=0;i<queue.size()-1;i++){
queue.add(queue.remove());
}

}

// Removes the element on top of the stack.
public void pop() {
queue.remove();
}

// Get the top element.
public int top() {
return queue.peek();
}

// Return whether the stack is empty.
public boolean empty() {
return queue.isEmpty();
}
}