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
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 Stack
{
public:
Stack(int = 10) ;
~Stack() { [] 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
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
解答： 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