lintcode: Implement Queue by Two Stacks

Description
Notes
Testcase
Judge

As the title described, you should only use two stacks to implement a queue's actions.
The queue should support

push(element)

,

pop()

and

top()

where
pop is pop the first(a.k.a front) element in the queue.
Both pop and top methods should return the value of first element.

Have you met this question in a real interview?

Yes

Example

push(1)
pop()     // return 1
push(2)
push(3)
top()     // return 2
pop()     // return 2

Challenge

implement it by two stacks, do not use any other data structure and push, pop and top should be O(1) by AVERAGE.

class Queue {
public:
stack<int> stack1;
stack<int> stack2;

Queue() {
// do intialization if necessary
}

void push(int element) {
// write your code here
while (stack2.size() > 0)
{
stack1.push(stack2.top());
stack2.pop();
}

stack1.push(element);
}

int pop() {

while (stack1.size() > 0)
{
stack2.push(stack1.top());
stack1.pop();
}

if (stack2.size() > 0)
{
int tmp = stack2.top();
stack2.pop();
return tmp;
}

return -1;
// write your code here
}

int top() {

while (stack1.size() > 0)
{
stack2.push(stack1.top());
stack1.pop();
}

if (stack2.size() > 0)
{
return stack2.top();
}

return -1;
// write your code here
}
};