Description#
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:
- You must use only standard operations of a queue, which means that only
push to back, peek/pop from front, size and is empty operations are valid. - Depending on your language, the queue may not be supported natively. You may simulate a queue using a list or deque (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 to push, pop, top, and empty. - All the calls to
pop and top are valid.
Follow-up: Can you implement the stack using only one queue?
Solutions#
Solution 1: Two Queues#
We use two queues $q_1$ and $q_2$, where $q_1$ is used to store the elements in the stack, and $q_2$ is used to assist in implementing the stack operations.
push operation: Push the element into $q_2$, then pop the elements in $q_1$ one by one and push them into $q_2$, finally swap the references of $q_1$ and $q_2$. The time complexity is $O(n)$.pop operation: Directly pop the front element of $q_1$. The time complexity is $O(1)$.top operation: Directly return the front element of $q_1$. The time complexity is $O(1)$.empty operation: Check whether $q_1$ is empty. The time complexity is $O(1)$.
The space complexity is $O(n)$, where $n$ is the number of elements in the stack.
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| class MyStack:
def __init__(self):
self.q1 = deque()
self.q2 = deque()
def push(self, x: int) -> None:
self.q2.append(x)
while self.q1:
self.q2.append(self.q1.popleft())
self.q1, self.q2 = self.q2, self.q1
def pop(self) -> int:
return self.q1.popleft()
def top(self) -> int:
return self.q1[0]
def empty(self) -> bool:
return len(self.q1) == 0
# Your MyStack object will be instantiated and called as such:
# obj = MyStack()
# obj.push(x)
# param_2 = obj.pop()
# param_3 = obj.top()
# param_4 = obj.empty()
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| import java.util.Deque;
class MyStack {
private Deque<Integer> q1 = new ArrayDeque<>();
private Deque<Integer> q2 = new ArrayDeque<>();
public MyStack() {
}
public void push(int x) {
q2.offer(x);
while (!q1.isEmpty()) {
q2.offer(q1.poll());
}
Deque<Integer> q = q1;
q1 = q2;
q2 = q;
}
public int pop() {
return q1.poll();
}
public int top() {
return q1.peek();
}
public boolean empty() {
return q1.isEmpty();
}
}
/**
* Your MyStack object will be instantiated and called as such:
* MyStack obj = new MyStack();
* obj.push(x);
* int param_2 = obj.pop();
* int param_3 = obj.top();
* boolean param_4 = obj.empty();
*/
|
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| class MyStack {
public:
MyStack() {
}
void push(int x) {
q2.push(x);
while (!q1.empty()) {
q2.push(q1.front());
q1.pop();
}
swap(q1, q2);
}
int pop() {
int x = q1.front();
q1.pop();
return x;
}
int top() {
return q1.front();
}
bool empty() {
return q1.empty();
}
private:
queue<int> q1;
queue<int> q2;
};
/**
* Your MyStack object will be instantiated and called as such:
* MyStack* obj = new MyStack();
* obj->push(x);
* int param_2 = obj->pop();
* int param_3 = obj->top();
* bool param_4 = obj->empty();
*/
|
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| type MyStack struct {
q1 []int
q2 []int
}
func Constructor() MyStack {
return MyStack{}
}
func (this *MyStack) Push(x int) {
this.q2 = append(this.q2, x)
for len(this.q1) > 0 {
this.q2 = append(this.q2, this.q1[0])
this.q1 = this.q1[1:]
}
this.q1, this.q2 = this.q2, this.q1
}
func (this *MyStack) Pop() int {
x := this.q1[0]
this.q1 = this.q1[1:]
return x
}
func (this *MyStack) Top() int {
return this.q1[0]
}
func (this *MyStack) Empty() bool {
return len(this.q1) == 0
}
/**
* Your MyStack object will be instantiated and called as such:
* obj := Constructor();
* obj.Push(x);
* param_2 := obj.Pop();
* param_3 := obj.Top();
* param_4 := obj.Empty();
*/
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| class MyStack {
q1: number[] = [];
q2: number[] = [];
constructor() {}
push(x: number): void {
this.q2.push(x);
while (this.q1.length) {
this.q2.push(this.q1.shift()!);
}
[this.q1, this.q2] = [this.q2, this.q1];
}
pop(): number {
return this.q1.shift()!;
}
top(): number {
return this.q1[0];
}
empty(): boolean {
return this.q1.length === 0;
}
}
/**
* Your MyStack object will be instantiated and called as such:
* var obj = new MyStack()
* obj.push(x)
* var param_2 = obj.pop()
* var param_3 = obj.top()
* var param_4 = obj.empty()
*/
|
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| use std::collections::VecDeque;
struct MyStack {
/// There could only be two status at all time
/// 1. One contains N elements, the other is empty
/// 2. One contains N - 1 elements, the other contains exactly 1 element
q_1: VecDeque<i32>,
q_2: VecDeque<i32>,
// Either 1 or 2, originally begins from 1
index: i32,
}
impl MyStack {
fn new() -> Self {
Self {
q_1: VecDeque::new(),
q_2: VecDeque::new(),
index: 1,
}
}
fn move_data(&mut self) {
// Always move from q1 to q2
assert!(self.q_2.len() == 1);
while !self.q_1.is_empty() {
self.q_2.push_back(self.q_1.pop_front().unwrap());
}
let tmp = self.q_1.clone();
self.q_1 = self.q_2.clone();
self.q_2 = tmp;
}
fn push(&mut self, x: i32) {
self.q_2.push_back(x);
self.move_data();
}
fn pop(&mut self) -> i32 {
self.q_1.pop_front().unwrap()
}
fn top(&mut self) -> i32 {
*self.q_1.front().unwrap()
}
fn empty(&self) -> bool {
self.q_1.is_empty()
}
}
|
