###### tags: `leetcode` `easy` `Stack` `Design` `Queue`
# [232. Implement Queue using Stacks](https://leetcode.com/problems/implement-queue-using-stacks/description/)

## 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.

## Examples
### Example1
**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:

- $1 \leq x \leq 9$
- 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](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.

## Code

typedef struct node {
    int data;
    struct node* next;
} NODE;

typedef struct {
    NODE *head, *tail;
} MyQueue;

MyQueue* myQueueCreate() {
    struct MyQueue* Q = calloc(1, sizeof(MyQueue));
    return Q;
}

void myQueuePush(MyQueue* obj, int x) {
    NODE* tmp = calloc(1, sizeof(NODE));
    tmp->data = x;
    tmp->next = NULL;
    if (!obj->head)
        obj->head = tmp;
    if (!obj->tail)
        obj->tail = tmp;
    else {
        obj->tail->next = tmp;
        obj->tail = obj->tail->next;
    }
}

int myQueuePop(MyQueue* obj) {
    NODE* tmp = obj->head;
    if (!tmp)
        return NULL;
    int data = tmp->data;
    if (obj->head == obj->tail) {
        free(tmp);
        obj->head = NULL;
        obj->tail = NULL;
    } else {
        obj->head = tmp->next;
        free(tmp);
    }
    return data;
}

int myQueuePeek(MyQueue* obj) {
    return obj->head->data;
}

bool myQueueEmpty(MyQueue* obj) {
    return obj->head == NULL;
}

void myQueueFree(MyQueue* obj) {
    NODE *pre, *tmp = obj->head;
    while (tmp != NULL) {
        pre = tmp;
        tmp = tmp->next;
        free(pre);
    }
}

## Complexity

|Space |Time |
|- |- |
|$O(N)$|$O(1)$|

## Result

- Runtime: 0 ms, 100%
- Memory: 5.9 MB, 59.26%