232. Implement Queue using Stacks

LeetCode 232. Implement Queue using Stacks

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.

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 to push, pop, peek, and empty.

• All the calls to pop and peek are valid.

Tags

Stack, Design

Solution

We use 2 stacks a and b, for receive and emit elements, respectively.

• Push: push x into a;

• Peek: When b is empty, pop all elements from a and push them into b, then return the top of b;

• Push: do Peek first, then pop the top of b;

• Empty: return true if no elements exist in both a and b.

Complexity

• Time complexity: $O(n)$ for Pop and Peek; $O(1)$ for others;

• Space complexity: $O(n)$

Code

type MyQueue struct {
	a, b []int
}

// Constructor initialize your data structure here.
func Constructor() MyQueue {
	return MyQueue{make([]int, 0, 100), make([]int, 0, 100)}
}

// Push element x to the back of queue.
func (this *MyQueue) Push(x int) {
	this.a = append(this.a, x)
}

// Pop removes the element from in front of queue and returns that element.
func (this *MyQueue) Pop() int {
	val := this.Peek()
	this.b = this.b[:len(this.b)-1]
	return val
}

// Peek gets the front element.
func (this *MyQueue) Peek() int {
	if len(this.b) == 0 {
		for len(this.a) > 0 {
			this.b = append(this.b, this.a[len(this.a)-1])
			this.a = this.a[:len(this.a)-1]
		}
	}
	return this.b[len(this.b)-1]
}

// Empty returns whether the queue is empty.
func (this *MyQueue) Empty() bool {
	return len(this.a) == 0 && len(this.b) == 0
}