华师大数据结构期中考试试卷(含答案).doc

. .华东师X大学期中试卷20072008学年第二学期课程名称：_数据结构_姓 名：_ 学 号：_专 业：_ 年级/班级：_课程性质：专业必修一二三四五六七八总分阅卷人签名一、 单项选择题(共18分，每题3分)1. Stack has the property called last in and first out, then which of the following describes the property of Queue?a) Last in and first outb) First in and last outc) First in and first out2. A list of items from which only the item most recently added can be removed is known as a ( )a) stack b) queue c) circular linked list d) list3. If the following function is called with a value of 2 for n, what is the resulting output?void Quiz( int n ) if (n > 0) cout << 0; Quiz(n - 1); cout << 1; Quiz(n - 1); a)00011011 b)11100100 c)10011100 d)01100011 e)0011014. A heap is a list in which each entry contains a key, and, for all positions i in the list, the key at position i is at lease as large as the keys in positions 2i+2 and ( ), provided these positions exist in the list.a) 2ib) 2i-1c) 2i-2d) 2i+15. Given the recursive function int Func( /* in */ int i, /* in */ int j ) if (i < 11) if (j < 11) return i + j; else return j + Func(i, j - 2); else return i + Func(i - 1, j); what is the value of the expression Func(12, 15) ? a)81 b)62 c)19 d)72 e)none of the above6. Shell sort finally perform an ordinary ( )?a) Heap sortb) Insertion sortc) Selection sortd) Quick sort二、 填空题（共22分，每空2分）1. If the following function is called with a value of 75 for n, the resulting output is _【1】_. void Func( /* in */ int n ) if (n > 0) Func(n / 8); cout << n % 8; 2. Give the output of the following program. _【2】_.template <class List_entry>void print(List_entry &x)cout<<x<<" "void main( )List<int> mylist;for(int i=0;i<5;i+)mylist.insert(i,i);cout<<"Your list have "<<mylist.size()<<" elements:"<<endl;mylist.remove(0,i);mylist.remove(2,i);mylist.insert(i,i);mylist.traverse(print);mylist.clear( );for(i=1;i<3;i+)mylist.insert(i, i);mylist.traverse(print);3. Read the following program and fill the blank to plete the method.template <class Node_entry>struct Node / data membersNode_entry entry;Node<Node_entry> *next;Node<Node_entry> *back;/ constructorsNode( );Node(Node_entry item, Node<Node_entry> *link_back = NULL, Node<Node_entry> *link_next = NULL);template <class List_entry>void List<List_entry> : set_position(int position) const/* Pre: position is a valid position in the List : 0 <=position < count .Post: The current Node pointer references the Node at position . */if (current_position <= position)for ( ; current_position != position; current_position+) 【3】 ;elsefor ( ; current_position != position; 【4】 )current = current->back;4. Read the following program and fill the blank to plete the method.Error_code recursive_binary_2(const Ordered_list &the_list, const Key &target, int bottom, int top, int &position)/* Pre: The indices bottom to top define the range in the list to search for the target .Post: If a Record in the range from bottom to top in the list has key equal to target , then position locates one such entry, and a code of success is returned. Otherwise, not_present is returned, and position is undefined.Uses: recursive_binary_2, together with methods from the classes Ordered_list and Record . */Record data;if (bottom <= top) int mid = 【5】 ;the_list.retrieve (mid, data);if (data = target) 【6】 ;return success;else if (data < target)return recursive_binary_2(the_list, target, 【7】 , top, position);elsereturn recursive_binary_2(the_list, target, bottom, 【8】 , position);else return not_present;5. The following program is the divide_from function of Merge Sort. Please fill the blank to plete the function. Node<Record> * divide_from (Node<Record> *sub_list)/* Post: The list of nodes referenced by sub_list has been reduced to its first half, and a pointer to the first node in the second half of the sublist is returned. If the sublist has an odd number of entries, then its first half will be one entry larger than its second.*/Node<Record> *position, / traverses the entire list*midpoint, / moves at half speed of position to midpoint*second_half;if (midpoint = sub_list) = NULL) return NULL; / List is empty.position = midpoint->next;while (【9】 ) / Move position twice for midpoint's one move.position = position->next;if (position != NULL) 【10】 ;position = position->next; second_half = 【11】 ;midpoint->next = NULL;return second_half;三、 编程题（共60分）1. （16分）Apply quicksort to the following list of 14 names, where the pivot in each sublist is chosen to be (a) the first key in the sublist and (b) the last key in the sublist. In each case, draw the tree of recursive call.Tim Dot Eva Roy Tom Kim Guy Amy Jon Ann Jim Kay Ron Jan2. （16分）Write the following overload operator for stacks:1) bool Stack:operator = (const Stack & s);2) bool Stack:operator += (const Stack & s);/ pushes the contents of the given stack onto this stack;3. （10分）Write the following function temple:Template <class T> void reverse(Queue <T> & q);/ reverses the contents of the given queue;4. （18分）struct Node / data membersint entry;Node *next;/ constructorsNode( );Node(int item, Node *link = NULL);f is the head pointer of a link list of nodes.Using recursion, implement the following functions:1) int Max (Node *f ); /return the max value in the link list.2) int Num (Node *f );/return the number of the nodes in the link list3) Node * Search ( Node *f, int x );/Search the first occurrence of the x in the link list. If success returns the /pointer to the node, else returns NULL.数据结构期中考卷参考答案. .word. .一、 单项选择题(3×618)1 c2 a3 e4 d5 a6 b二、 填空题(2×1122)【1】113【2】Your list have 5 elements:1 2 4 3【3】current = current->next【4】current_position-【5】(bottom + top)/2【6】position = mid【7】mid + 1【8】mid 1【9】position != NULL【10】midpoint = midpoint->next;【11】midpoint->next. .word. . .word. .三、 编程题(16×2101860)1，a) P344 F8.12b)2. 1) bool Stack:operator=(const Stack &s)Stack s1=s, s2=*this;while (!s1.empty( )if (s1.top( )!= s2.top( ) return false;else s1.pop( ); s2.pop( );2) bool Stack: operator+=(const Stack &s)Stack ss=s, s2;while (!ss.empty( )s2.push(ss.top( );ss.pop( );while (!s2.empty( )if(push(s2.top( ) = oute) return false;s2.pop( );3Template <class T> void reverse (Queue<T> & q)Stack<T> s  T data;while ( !q.empty( ) q.retrieve(data ); s.push( data); q.serve( );while (!s.empty( )q.append(s.top( );s.pop( );4. int Max (Node *f ) if ( f ->next = NULL ) return f ->entry;int temp = Max ( f ->next );if ( f ->data > temp ) return f ->entry;else return temp;int Num ( Node *f ) if ( f = NULL ) return 0;return 1+ Num ( f ->next );Node * Search ( Node *f, int x ) if ( f = NULL)return NULL;else if ( f entry = x ) return f; else return Search( fnext , x);. .word.