Description
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:
- You must use only standard operations of a stack, which means only
push to top, peek/pop from top, size, and is 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.
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 to push, pop, peek, and empty.
- All the calls to
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.
Solutions
Solution 1: Double Stack
We use two stacks, where stk1 is used for enqueue, and another stack stk2 is used for dequeue.
When enqueueing, we directly push the element into stk1. The time complexity is O(1).
When dequeueing, we first check whether stk2 is empty. If it is empty, we pop all elements from stk1 and push them into stk2, and then pop an element from stk2. If stk2 is not empty, we directly pop an element from stk2. The amortized time complexity is O(1).
When getting the front element, we first check whether stk2 is empty. If it is empty, we pop all elements from stk1 and push them into stk2, and then get the top element from stk2. If stk2 is not empty, we directly get the top element from stk2. The amortized time complexity is O(1).
When checking whether the queue is empty, we only need to check whether both stacks are empty. The time complexity is O(1).
PythonJavaC++GoTypeScriptRust
class MyQueue:
def __init__(self):
self.stk1 = []
self.stk2 = []
def push(self, x: int) -> None:
self.stk1.append(x)
def pop(self) -> int:
self.move()
return self.stk2.pop()
def peek(self) -> int:
self.move()
return self.stk2[-1]
def empty(self) -> bool:
return not self.stk1 and not self.stk2
def move(self):
if not self.stk2:
while self.stk1:
self.stk2.append(self.stk1.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()(code-box)
class MyQueue {
private Deque<Integer> stk1 = new ArrayDeque<>();
private Deque<Integer> stk2 = new ArrayDeque<>();
public MyQueue() {
}
public void push(int x) {
stk1.push(x);
}
public int pop() {
move();
return stk2.pop();
}
public int peek() {
move();
return stk2.peek();
}
public boolean empty() {
return stk1.isEmpty() && stk2.isEmpty();
}
private void move() {
while (stk2.isEmpty()) {
while (!stk1.isEmpty()) {
stk2.push(stk1.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();
*/(code-box)
class MyQueue {
public:
MyQueue() {
}
void push(int x) {
stk1.push(x);
}
int pop() {
move();
int ans = stk2.top();
stk2.pop();
return ans;
}
int peek() {
move();
return stk2.top();
}
bool empty() {
return stk1.empty() && stk2.empty();
}
private:
stack<int> stk1;
stack<int> stk2;
void move() {
if (stk2.empty()) {
while (!stk1.empty()) {
stk2.push(stk1.top());
stk1.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();
* bool param_4 = obj->empty();
*/(code-box)
type MyQueue struct {
stk1 []int
stk2 []int
}
func Constructor() MyQueue {
return MyQueue{[]int{}, []int{}}
}
func (this *MyQueue) Push(x int) {
this.stk1 = append(this.stk1, x)
}
func (this *MyQueue) Pop() int {
this.move()
ans := this.stk2[len(this.stk2)-1]
this.stk2 = this.stk2[:len(this.stk2)-1]
return ans
}
func (this *MyQueue) Peek() int {
this.move()
return this.stk2[len(this.stk2)-1]
}
func (this *MyQueue) Empty() bool {
return len(this.stk1) == 0 && len(this.stk2) == 0
}
func (this *MyQueue) move() {
if len(this.stk2) == 0 {
for len(this.stk1) > 0 {
this.stk2 = append(this.stk2, this.stk1[len(this.stk1)-1])
this.stk1 = this.stk1[:len(this.stk1)-1]
}
}
}
/**
* Your MyQueue object will be instantiated and called as such:
* obj := Constructor();
* obj.Push(x);
* param_2 := obj.Pop();
* param_3 := obj.Peek();
* param_4 := obj.Empty();
*/(code-box)
class MyQueue {
stk1: number[];
stk2: number[];
constructor() {
this.stk1 = [];
this.stk2 = [];
}
push(x: number): void {
this.stk1.push(x);
}
pop(): number {
this.move();
return this.stk2.pop();
}
peek(): number {
this.move();
return this.stk2.at(-1);
}
empty(): boolean {
return !this.stk1.length && !this.stk2.length;
}
move(): void {
if (!this.stk2.length) {
while (this.stk1.length) {
this.stk2.push(this.stk1.pop()!);
}
}
}
}
/**
* 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()
*/(code-box)
use std::collections::VecDeque;
struct MyQueue {
stk1: Vec<i32>,
stk2: Vec<i32>,
}
impl MyQueue {
fn new() -> Self {
MyQueue {
stk1: Vec::new(),
stk2: Vec::new(),
}
}
fn push(&mut self, x: i32) {
self.stk1.push(x);
}
fn pop(&mut self) -> i32 {
self.move_elements();
self.stk2.pop().unwrap()
}
fn peek(&mut self) -> i32 {
self.move_elements();
*self.stk2.last().unwrap()
}
fn empty(&self) -> bool {
self.stk1.is_empty() && self.stk2.is_empty()
}
fn move_elements(&mut self) {
if self.stk2.is_empty() {
while let Some(element) = self.stk1.pop() {
self.stk2.push(element);
}
}
}
}(code-box)
