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# positive grid 面試筆試題目

### 筆試

#### 第一題
考 LeetCode 155 min stack
計時 10 分鐘

寫白板模擬花了 8 分鐘
約 3 分鐘把 pseudo code 寫完

然後一行一行 code review
應該沒什麼錯誤

時間複雜度 O(1) (for 每個操作)
空間複雜度 O(1)

type MinStack struct {
    
}


/** initialize your data structure here. */
func Constructor() MinStack {
    
}


func (this *MinStack) Push(x int)  {
    
}


func (this *MinStack) Pop()  {
    
}


func (this *MinStack) Top() int {
    
}


func (this *MinStack) GetMin() int {
    
}


/**
 * Your MinStack object will be instantiated and called as such:
 * obj := Constructor();
 * obj.Push(x);
 * obj.Pop();
 * param_3 := obj.Top();
 * param_4 := obj.GetMin();
 */

其實看到題目當下我就直接說了這題我寫過R
是想看看對方會不會換題目，不過他們說寫過也沒差，先寫出來。

用差值的方式解完之後對方說這個方法比較 trick ，第一次看到應該不太可能想出來
我們就老實探討如果第一次看到用 O(1) O(n) 的方式解的話怎樣解

用一個陣列可以做到 O(logn) O(n)，用兩個陣列模擬狀態機可以做到 O(1) O(n)

#### 第二題

# include <iostream>

using namespace std;
class base {
    public:
        base () {
            cout << "base controctor" << endl;
        }
        ~base () {
            cout << "base detroctor" << endl;
        }
};

class derived : public base {
    public:
        derived () {
            cout << "derived controctor" << endl;
        }
        ~derived (){
            cout << "derived detroctor" << endl;
        }
};

int main() {
    base* a = new derived();
    delete a;
}

問輸出

實際輸出應該是

base controctor
derived controctor
base detroctor

說有沒有辦法呼叫到 `cout << "derived detroctor" << endl;`

加個 virtaul 就好

    virtual ~base () {
        cout << "base detroctor" << endl;
    }

#### 第三題

# include <iostream>

using namespace std;
class base {
    public:
        virtual void print(){
            cout << "derived print" <<endl;
        }
        void show(){
            cout << "derived show" <<endl;
        }
};

class derived : public base {
    public:
        void print(){
            cout << "derived print" <<endl;
        }
        void show(){
            cout << "derived show" <<endl;
        }
};

int main() {
    base* a = new derived();
    a -> print();
    a -> show();
    derived d;
    base* c = &d;
    c -> print();
    c -> show();
    return 0;
}

問輸出
實際答案

derived print
base show
derived print
base show

順帶一提

    base* a = new derived();
    // and
    derived d;
    base* c = &d;

兩者是等價的

口試

聊聊
vector map
RAII
Design Patterns
dynamic_cast static_cast
circular reference
how to debug

後面跟 manager 還有聊到

git flow
線程模型