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()
Returnstrue
if the queue is empty,false
otherwise.
Notes:
- You must use only standard operations of a stack, which means only
push to top
,peek/pop from top
,size
, andis empty
operations are valid. - Depending on your language, the stack may not be supported natively. You may simulate a stack using a list or deque (double-ended queue) as long as you use only a stack's standard operations.
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.
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
- At most
100
calls will be made topush
,pop
,peek
, andempty
. - All the calls to
pop
andpeek
are valid.
class MyQueue:
def __init__(self):
"""
Initialize your data structure here.
"""
self.s1 = []
self.s2 = []
def push(self, x: int) -> None:
"""
Push element x to the back of queue.
"""
self.s1.append(x)
def pop(self) -> int:
"""
Removes the element from in front of queue and returns that element.
"""
self._move()
return self.s2.pop()
def peek(self) -> int:
"""
Get the front element.
"""
self._move()
return self.s2[-1]
def empty(self) -> bool:
"""
Returns whether the queue is empty.
"""
return len(self.s1) + len(self.s2) == 0
def _move(self):
"""
Move elements from s1 to s2.
"""
if len(self.s2) == 0:
while len(self.s1) > 0:
self.s2.append(self.s1.pop())
# Your MyQueue object will be instantiated and called as such:
# obj = MyQueue()
# obj.push(x)
# param_2 = obj.pop()
# param_3 = obj.peek()
# param_4 = obj.empty()
class MyQueue {
private Deque<Integer> s1 = new ArrayDeque<>();
private Deque<Integer> s2 = new ArrayDeque<>();
/** Initialize your data structure here. */
public MyQueue() {
}
/** Push element x to the back of queue. */
public void push(int x) {
s1.push(x);
}
/** Removes the element from in front of queue and returns that element. */
public int pop() {
move();
return s2.pop();
}
/** Get the front element. */
public int peek() {
move();
return s2.peek();
}
/** Returns whether the queue is empty. */
public boolean empty() {
return s1.isEmpty() && s2.isEmpty();
}
/** Move elements from s1 to s2. */
private void move() {
if (s2.isEmpty()) {
while (!s1.isEmpty()) {
s2.push(s1.pop());
}
}
}
}
/**
* Your MyQueue object will be instantiated and called as such:
* MyQueue obj = new MyQueue();
* obj.push(x);
* int param_2 = obj.pop();
* int param_3 = obj.peek();
* boolean param_4 = obj.empty();
*/
class MyQueue {
stack1: number[];
stack2: number[];
constructor() {
this.stack1 = [];
this.stack2 = [];
}
push(x: number): void {
this.stack1.push(x);
}
pop(): number {
if (!this.stack2.length) {
while (this.stack1.length) {
this.stack2.push(this.stack1.pop());
}
}
return this.stack2.pop();
}
peek(): number {
if (!this.stack2.length) {
while (this.stack1.length) {
this.stack2.push(this.stack1.pop());
}
}
return this.stack2[this.stack2.length - 1];
}
empty(): boolean {
return !this.stack1.length && !this.stack2.length;
}
}
/**
* Your MyQueue object will be instantiated and called as such:
* var obj = new MyQueue()
* obj.push(x)
* var param_2 = obj.pop()
* var param_3 = obj.peek()
* var param_4 = obj.empty()
*/