Easy
Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (push
, peek
, pop
, and empty
).
Implement the MyQueue
class:
void push(int x)
Pushes element x to the back of the queue.int pop()
Removes the element from the front of the queue and returns it.int peek()
Returns the element at the front of the queue.boolean empty()
Returns true
if the queue is empty, false
otherwise.Notes:
push to top
, peek/pop from top
, size
, and is empty
operations are valid.Example 1:
Input:
["MyQueue", "push", "push", "peek", "pop", "empty"]
[[], [1], [2], [], [], []]
Output:
[null, null, null, 1, 1, false]
Explanation:
MyQueue myQueue = new MyQueue();
myQueue.push(1); // queue is: [1]
myQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue)
myQueue.peek(); // return 1
myQueue.pop(); // return 1, queue is [2]
myQueue.empty(); // return false
Constraints:
1 <= x <= 9
100
calls will be made to push
, pop
, peek
, and empty
.pop
and peek
are valid.Follow-up: Can you implement the queue such that each operation is amortized O(1)
time complexity? In other words, performing n
operations will take overall O(n)
time even if one of those operations may take longer.
import java.util.Deque
import java.util.LinkedList
class MyQueue() {
private val deque: Deque<Int> = LinkedList()
fun push(x: Int) {
deque.add(x)
}
fun pop(): Int = deque.removeFirst()
fun peek(): Int = deque.first()
fun empty(): Boolean = deque.isEmpty()
}
/*
* Your MyQueue object will be instantiated and called as such:
* var obj = MyQueue()
* obj.push(x)
* var param_2 = obj.pop()
* var param_3 = obj.peek()
* var param_4 = obj.empty()
*/