一些笔试题目和整理的答案 - 腾讯(Tencent)

aftyingjiesheng · 发表于 2012-3-28 17:45

一些笔试题目和整理的答案 - 腾讯(Tencent)

NO1
Below is usual way we find one element in an array
const int *find1(const int* array, int n, int x)
{
const int* p = array;
for(int i = 0; i < n; i++)
{
      if(*p == x)
      {
         return p;
      }
      ++p;
}
return 0; }
In this case we have to bear the knowledge of value type "int", the size of array, even the existence of an array. Would you re-write it using template to eliminate all these dependencies?

template <class T>
const T *find1(const T* array, int n, T x)
{
const T* p = array;
for(int i = 0; i < n; i++)
{
      if(*p == x)
      {
         return p;
      }
      ++p;
}
return 0; }

NO2

Give an example of implementing a Stack in the template way(only template class declaration without detail definition and realization)
template <class T>
class Stack
{
public:
   Stack(int = 10) ;
   ~Stack() { delete [] stackPtr ; }
   int push(const T&);
   int pop(T&) ;
   int isEmpty()const { return top == -1 ; }
   int isFull() const { return top == size - 1 ; }
private:
   int size ;  // number of elements on Stack.
   int top ;
   T* stackPtr ;
} ;

NO3

Implement the simplest singleton pattern(initialize if necessary).
class Singleton {
public:
static Singleton* Instance();
protected:
Singleton();
private:
static Singleton* _instance;
}

// Implementation
Singleton* Singleton::_instance = 0;

Singleton* Singleton::Instance() {
if (_instance == 0) {
      _instance = new Singleton;
}
return _instance;
}

NO4

1.Jeff and Diamond like playing game of coins, One day they designed a new set of rules:
1)Totally 10 coins
2)One can take away 1,2or 4 coins at one time by turns
3)Who takes the last loses.
Given these rules Whether the winning status is pre-determined or not

1：从后面开始考虑，最后肯定要留1个才能保证自己赢
2：所以要设法让对方留下2，3，5个
3：也就是要自己取后留下1，4，6，7，8，9
4：如果自己取后留下6，对方取2个，与（3）矛盾，所以排除6
5：如果自己取后留下8，对方取4个，与（3）一样情况，所以也排除8
6：同样，9也不行，如果我抽后剩下9，对方抽2个，就反过来成对方抽剩成7个了，也与3）矛盾，所以也排除
7：所以很显然，我只能抽剩1，4，7
8：因为只能抽后剩1，4，7才能赢，我先抽得话不可能达到这几个数，很显然，只能让对
方先抽，也即是先抽的人输

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