Question

Formatted question description: https://leetcode.ca/all/232.html

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.

Algorithm

As long as the element is inserted from the front every time, for example, if a queue is 1, 2, 3, 4, then we save it in the stack as 4, 3, 2, 1, and then return to the top element 1, also It is the first element of the queue.

The difficulty of this question is the push function. We need an auxiliary stack tmp, and store the elements of s into tmp in reverse order. At this time, add a new element x, and then store the elements in tmp back. This is the order we want , The other three operations also directly call stack operations.

To optimize above, because every time you push, you have to flip both stacks,

So, have stacks stk1 and stk2, the new pushed elements are cached in stk1 first, and all elements in stk1 are moved to stk2 only when the operations are pop and peek, which improves efficiency.

Code

• 
public class Implement_Queue_using_Stacks {
class MyQueue {
private Stack<Integer> stack1;
private Stack<Integer> stack2;

public MyQueue() {
this.stack1 = new Stack<Integer>();
this.stack2 = new Stack<Integer>();
}
// Push element x to the back of queue.
public void push(int x) {
stack1.push(x);
}

// Removes the element from in front of queue.
public int pop() {
if (!stack2.isEmpty()) {
return stack2.pop(); // stack is queue-order, queue-head at this-stack-top
} else {
while (!stack1.isEmpty()) {
stack2.push(stack1.pop());
}
return stack2.pop();
}
}

// Get the front element.
public int peek() {
int ret = 0;
if (!stack2.isEmpty()) {
ret = stack2.peek();
} else {
while (!stack1.isEmpty()) {
stack2.push(stack1.pop());
}
ret = stack2.peek();
}

return ret;
}

// Return whether the queue is empty.
public boolean empty() {
return stack1.isEmpty() && stack2.isEmpty();
}
}
}

############

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();
*/

• // OJ: https://leetcode.com/problems/implement-queue-using-stacks
// Time: O(1) amortized.
// Space: O(1)
class MyQueue {
stack<int> in, out;
public:
MyQueue() {}

void push(int x) {
in.push(x);
}

int pop() {
int val = peek();
out.pop();
return val;
}

int peek() {
if (out.empty()) {
while (in.size()) {
out.push(in.top());
in.pop();
}
}
return out.top();
}

bool empty() {
return in.empty() && out.empty();
}
};

• class MyQueue:
def __init__(self):
self.stk1 = []
self.stk2 = [] # reversed order

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: # only when skt2 is empty
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()

############

class MyQueue:

def __init__(self):
self.sk = []
self.rsk = [] # reversed

def push(self, x: int) -> None:
self.sk.append(x);

def pop(self) -> int:
self.peek()
return self.rsk.pop()

def peek(self) -> int:
if self.rsk:
return self.rsk[-1]
else:
while self.sk:
self.rsk.append(self.sk.pop())
return self.rsk[-1]

def empty(self) -> bool:

return not (self.sk or self.rsk)

# 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()

• 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();
*/

• 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[this.stk2.length - 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()
*/


• struct MyQueue {
in_stack: Vec<i32>,
out_stack: Vec<i32>,
}

/**
* &self means the method takes an immutable reference.
* If you need a mutable reference, change it to &mut self instead.
*/
impl MyQueue {

fn new() -> Self {
Self {
in_stack: vec![],
out_stack: vec![],
}
}

fn push(&mut self, x: i32) {
self.in_stack.push(x);
}

fn pop(&mut self) -> i32 {
if self.out_stack.is_empty() {
self.fill_out();
}
self.out_stack.pop().unwrap()
}

fn peek(&mut self) -> i32 {
if self.out_stack.is_empty() {
self.fill_out();
}
*self.out_stack.last().unwrap()
}

fn empty(&self) -> bool {
self.in_stack.is_empty() && self.out_stack.is_empty()
}

fn fill_out(&mut self){
let MyQueue { in_stack, out_stack } = self;
if out_stack.is_empty() {
while !in_stack.is_empty() {
out_stack.push(in_stack.pop().unwrap());
}
}
}
}

/**
* Your MyQueue object will be instantiated and called as such:
* let obj = MyQueue::new();
* obj.push(x);
* let ret_2: i32 = obj.pop();
* let ret_3: i32 = obj.peek();
* let ret_4: bool = obj.empty();
*/