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# 225. Implement Stack using Queues

## LeetCode [225. Implement Stack using Queues](https://github.com/txfs19260817/LeetCode-Go/blob/master/solutions/title/README.md)

### Description

Implement a last in first out (LIFO) stack using only two queues. The implemented stack should support all the functions of a normal queue (`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 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 such that each operation is [**amortized**](https://en.wikipedia.org/wiki/Amortized_analysis) `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. You can use more than two queues.

### Tags

Stack, Design

### Solution

Here we only use a single queue to implement the stack.

* Push: push `x` into the queue, then pop and push all other elements;
* Pop: same as the queue's pop;
* Top: return the value of the front element of the queue;
* Empty: return the equality of the length of queue and 0.

### Complexity

* Time complexity: $$O(n)$$ for Push; $$O(1)$$ for others;
* Space complexity: $$O(n)$$

### Code

```go
type MyStack struct {
	q []int
}

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

// Push element x onto stack. */
func (this *MyStack) Push(x int) {
	n := len(this.q)
	this.q = append(this.q, x)
	for i := 0; i < n; i++ {
		this.q = append(this.q, this.q[0])
		this.q = this.q[1:]
	}
}

// Pop removes the element on top of the stack and returns that element. */
func (this *MyStack) Pop() int {
	val := this.q[0]
	this.q = this.q[1:]
	return val
}

// Top get the top element. */
func (this *MyStack) Top() int {
	return this.q[0]
}

// Empty returns whether the stack is empty. */
func (this *MyStack) Empty() bool {
	return len(this.q) == 0
}
```

## Reference

1. [用队列实现栈](https://leetcode-cn.com/problems/implement-stack-using-queues/solution/yong-dui-lie-shi-xian-zhan-by-leetcode-solution/)