Easy
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 <= 9
100
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?
import java.util.LinkedList
import java.util.Queue
class MyStack() {
private val queue1: Queue<Int> = LinkedList()
private val queue2: Queue<Int> = LinkedList()
fun push(x: Int) {
queue1.add(x)
}
fun pop(): Int {
while (queue1.size > 1) {
queue2.add(queue1.remove())
}
val top = queue1.remove()
queue1.clear()
queue1.addAll(queue2)
queue2.clear()
return top
}
fun top(): Int {
while (queue1.size > 1) {
queue2.add(queue1.remove())
}
val top = queue1.remove()
queue2.add(top)
queue1.clear()
queue1.addAll(queue2)
queue2.clear()
return top
}
fun empty(): Boolean {
return queue1.isEmpty()
}
}
/*
* Your MyStack object will be instantiated and called as such:
* var obj = MyStack()
* obj.push(x)
* var param_2 = obj.pop()
* var param_3 = obj.top()
* var param_4 = obj.empty()
*/