---
title: 'LeetCode 225. Implement Stack using Queues'
disqus: hackmd
---

# LeetCode 225. Implement Stack using Queues

## Description

Implement a last-in-first-out (LIFO) stack using only two queues. The implemented stack should support all the functions of a normal stack (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 that 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

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.

## Answer

此題可用Queue的概念 (FIFO) 實現stack (LIFO)，比較需要注意的就是用兩個Queue就可模擬stack，在pop時就將Q1從頭塞到Q2直到最後一個，然後將最後一個pop出去，再將Q1 & Q2做SWAP即可，相當於拿Q2來暫存。

typedef struct {
    int *Q1;
    int *Q2;
    int max;
    int top;
} MyStack;


MyStack* myStackCreate() {
    MyStack *obj = (MyStack*)malloc(sizeof(MyStack));
    obj->max = 100;
    obj->top = 0;
    obj->Q1 = (int*)malloc(sizeof(int)*(obj->max));
    obj->Q2 = (int*)malloc(sizeof(int)*(obj->max));
    return obj;
}

void myStackPush(MyStack* obj, int x) {
     if(obj->top < obj->max){
         obj->Q1[obj->top] = x;
         (obj->top)++;
     }
}

int myStackPop(MyStack* obj) {
    int ans = 0, i = 0;  
    int *temp = NULL;
    if(obj->top > 0){
        for(i = 0; i < obj->top - 1; i++){obj->Q2[i] = obj->Q1[i];}
        ans = obj->Q1[i];
        (obj->top)--;
        temp = obj->Q1;
        obj->Q1 = obj->Q2;
        obj->Q2 = temp;
    }
    return ans;  
}

int myStackTop(MyStack* obj) {
    int ans = 0;
    if(obj->top > 0){
        ans = obj->Q1[obj->top - 1];
    }
    return ans; 
}

bool myStackEmpty(MyStack* obj) {
  return obj->top == 0;
}

void myStackFree(MyStack* obj) {
    free(obj->Q1);
    free(obj->Q2);
    free(obj);
}

## Link
https://leetcode.com/problems/implement-stack-using-queues/
###### tags: `Leetcode`