Implement a last-in-first-out (LIFO) stack using only two queues. The implemented stack should support all the functions of a normal stack (push, top, pop, and empty).
Implement the MyStack class:
void push(int x) Pushes element x to the top of the stack.int pop() Removes the element on the top of the stack and returns it.int top() Returns the element on the top of the stack.boolean empty() Returns true if the stack is empty, false otherwise.Notes:
push to back, peek/pop from front, size and is empty operations are valid.
Example 1:
Input ["MyStack", "push", "push", "top", "pop", "empty"] [[], [1], [2], [], [], []] Output [null, null, null, 2, 2, false] Explanation MyStack myStack = new MyStack(); myStack.push(1); myStack.push(2); myStack.top(); // return 2 myStack.pop(); // return 2 myStack.empty(); // return False
Constraints:
1 <= x <= 9100 calls will be made to push, pop, top, and empty.pop and top are valid.
Follow-up: Can you implement the stack using only one queue?
class MyStack {
Queue<Integer> q1 = new LinkedList<>();
Queue<Integer> q2 = new LinkedList<>();
public MyStack() {
}
public void push(int x) {
while (!q1.isEmpty())
q2.offer(q1.poll());
q1.offer(x);
while (!q2.isEmpty())
q1.offer(q2.poll());
}
public int pop() {
// if (q1.isEmpty()) // because constraints mention always valid
// return -1;
return q1.poll();
}
public int top() {
return q1.peek();
}
public boolean empty() {
return q1.isEmpty();
}
}
/**
* Your MyStack object will be instantiated and called as such:
* MyStack obj = new MyStack();
* obj.push(x);
* int param_2 = obj.pop();
* int param_3 = obj.top();
* boolean param_4 = obj.empty();
*/