225, Implement Stack using Queues
About 2 min
I Problem
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()
Returnstrue
if the stack is empty,false
otherwise.
Notes:
- You must use only standard operations of a queue, which means that only
push to back
,peek/pop from front
,size
andis empty
operations are valid. - Depending on your language, the queue may not be supported natively. You may simulate a queue using a
list
ordeque
(double-ended queue) as long as you use only a queue's standard operations.
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
- At most
100
calls will be made topush
,pop
,top
, andempty
. - All the calls to
pop
andtop
are valid.
Follow up
Can you implement the stack using only one queue?
Related Topics
- Stack
- Design
- Queue
II Solution
Approach 1: Two Queues
pub struct MyStack {
q1: VecDeque<i32>,
q2: VecDeque<i32>,
}
impl MyStack {
pub fn new() -> Self {
MyStack {
q1: VecDeque::new(),
q2: VecDeque::new(),
}
}
/// Time Complexity: O(n)
///
/// Space Complexity: O(n)
pub fn push(&mut self, x: i32) {
self.q2.push_back(x);
while let Some(val) = self.q1.pop_front() {
self.q2.push_back(val);
}
std::mem::swap(&mut self.q1, &mut self.q2);
}
/// Time Complexity: O(1)
///
/// Space Complexity: O(1)
pub fn pop(&mut self) -> i32 {
self.q1.pop_front().unwrap()
}
/// Time Complexity: O(1)
///
/// Space Complexity: O(1)
pub fn top(&mut self) -> i32 {
*self.q1.front().unwrap()
}
/// Time Complexity: O(1)
///
/// Space Complexity: O(1)
pub fn empty(&self) -> bool {
self.q1.is_empty()
}
}
public class MyStack {
private Deque<Integer> q1;
private Deque<Integer> q2;
public MyStack() {
this.q1 = new ArrayDeque<>();
this.q2 = new ArrayDeque<>();
}
// Time Complexity: O(n)
//
// Space Complexity: O(n)
public void push(int x) {
this.q2.add(x);
while (!this.q1.isEmpty()) {
this.q2.add(this.q1.remove());
}
Deque<Integer> temp = this.q1;
this.q1 = this.q2;
this.q2 = temp;
}
// Time Complexity: O(1)
//
// Space Complexity: O(1)
public int pop() {
return this.q1.remove();
}
// Time Complexity: O(1)
//
// Space Complexity: O(1)
public int top() {
return this.q1.peek();
}
// Time Complexity: O(1)
//
// Space Complexity: O(1)
public boolean empty() {
return this.q1.isEmpty();
}
}
Approach 2: One Queue
pub struct MyStack {
q1: VecDeque<i32>,
}
impl MyStack {
pub fn new() -> Self {
MyStack {
q1: VecDeque::new(),
}
}
/// Time Complexity: O(n)
///
/// Space Complexity: O(n)
pub fn push(&mut self, x: i32) {
let mut len = self.q1.len();
self.q1.push_back(x);
while len != 0 {
if let Some(val) = self.q1.pop_front() {
self.q1.push_back(val);
}
len -= 1;
}
}
/// Time Complexity: O(1)
///
/// Space Complexity: O(1)
pub fn pop(&mut self) -> i32 {
self.q1.pop_front().unwrap()
}
/// Time Complexity: O(1)
///
/// Space Complexity: O(1)
pub fn top(&mut self) -> i32 {
*self.q1.front().unwrap()
}
/// Time Complexity: O(1)
///
/// Space Complexity: O(1)
pub fn empty(&self) -> bool {
self.q1.is_empty()
}
}
public class MyStack {
private Deque<Integer> q1;
public MyStack() {
this.q1 = new ArrayDeque<>();
}
// Time Complexity: O(n)
//
// Space Complexity: O(n)
public void push(int x) {
int size = this.q1.size();
this.q1.add(x);
while (size-- != 0) {
this.q1.add(this.q1.remove());
}
}
// Time Complexity: O(1)
//
// Space Complexity: O(1)
public int pop() {
return this.q1.remove();
}
// Time Complexity: O(1)
//
// Space Complexity: O(1)
public int top() {
return this.q1.peek();
}
// Time Complexity: O(1)
//
// Space Complexity: O(1)
public boolean empty() {
return this.q1.isEmpty();
}
}