tags: `leetcode` `easy` `Stack` `Design` `Queue`

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.

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 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%

tags: leetcode easy Stack Design Queue