40. 链表题：一个链表的结点结构

struct Node

{

int data ;

Node *next ;

};

typedef struct Node Node ;

(1)已知链表的头结点head,写一个函数把这个链表逆序 ( Intel)

Node * ReverseList(Node *head) //链表逆序

{

if ( head == NULL || head->next == NULL )

return head;

Node *p1 = head ;

Node *p2 = p1->next ;

Node *p3 = p2->next ;

p1->next = NULL ;

while ( p3 != NULL )

{

p2->next = p1 ;

p1 = p2 ;

p2 = p3 ;

p3 = p3->next ;

}

p2->next = p1 ;

head = p2 ;

return head ;

}

(2)已知两个链表head1 和head2 各自有序，请把它们合并成一个链表依然有序。(保留所有结点，即便大小相同）

Node * Merge(Node *head1 , Node *head2)

{

if ( head1 == NULL)

return head2 ;

if ( head2 == NULL)

return head1 ;

Node *head = NULL ;

Node *p1 = NULL;

Node *p2 = NULL;

if ( head1->data < head2->data )

{

head = head1 ;

p1 = head1->next;

p2 = head2 ;

}

else

{

head = head2 ;

p2 = head2->next ;

p1 = head1 ;

}

Node *pcurrent = head ;

while ( p1 != NULL && p2 != NULL)

{

if ( p1->data <= p2->data )

{

pcurrent->next = p1 ;

pcurrent = p1 ;

p1 = p1->next ;

}

else

{

pcurrent->next = p2 ;

pcurrent = p2 ;

p2 = p2->next ;

}

}

if ( p1 != NULL )

pcurrent->next = p1 ;

if ( p2 != NULL )

pcurrent->next = p2 ;

return head ;

}

(3)已知两个链表head1 和head2 各自有序，请把它们合并成一个链表依然有序，这次要求用递归方法进行。 (Autodesk)

答案：

Node * MergeRecursive(Node *head1 , Node *head2)

{

if ( head1 == NULL )

return head2 ;

if ( head2 == NULL)

return head1 ;

Node *head = NULL ;

if ( head1->data < head2->data )

{

head = head1 ;

head->next = MergeRecursive(head1->next,head2);

}

else

{

head = head2 ;

head->next = MergeRecursive(head1,head2->next);

}

return head ;

}

41. 分析一下这段程序的输出 (Autodesk)

class B

{

public:

B()

{

cout<<"default constructor"<<endl;

}

~B()

{

cout<<"destructed"<<endl;

}

B(int i):data(i) //B(int) works as a converter ( int -> instance of B)

{

cout<<"constructed by parameter " << data <<endl;

}

private:

int data;

};

B Play( B b)

{

return b ;

}

(1) results:

int main(int argc, char* argv[]) constructed by parameter 5

{ destructed B(5)形参析构

B t1 = Play(5); B t2 = Play(t1); 　 destructed t1形参析构

return 0;　　　　　　　　　　　　　　 destructed t2　注意顺序！

} destructed t1

(2) results:

int main(int argc, char* argv[]) constructed by parameter 5

{ destructed B(5)形参析构

B t1 = Play(5); B t2 = Play(10); 　 constructed by parameter 10

return 0;　　　　　　　　　　　　　　 destructed B(10)形参析构

} destructed t2　注意顺序！

destructed t1

42. 写一个函数找出一个整数数组中，第二大的数 （microsoft）

答案：

const int MINNUMBER = -32767 ;

int find_sec_max( int data[] , int count)

{

int maxnumber = data[0] ;

int sec_max = MINNUMBER ;

for ( int i = 1 ; i < count ; i++)

{

if ( data[i] > maxnumber )

{

sec_max = maxnumber ;

maxnumber = data[i] ;

}

else

{

if ( data[i] > sec_max )

sec_max = data[i] ;

}

}

return sec_max ;

}