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数据结构基础 期中考试解答. 2011-10-8. 选择 1. Which of the following is true? a. n ! grows faster than x n for any x b. ( log n) 2 grows faster than n 1/2 c. n grows slower than n 1/2 (log n) 2 d. log(n 2 ) grows faster than (log n) 2 解答：选 a 增长速度： N ！ > 指数型 > 多项式型 > 对数型. 选择 2.
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数据结构基础期中考试解答 2011-10-8
选择1 • Which of the following is true? • a. n! grows faster than xnfor any x • b. (log n)2 grows faster than n1/2 • c. n grows slower than n1/2(log n)2 • d. log(n2) grows faster than (log n)2 • 解答：选a • 增长速度： • N！>指数型>多项式型>对数型
选择2 • Two stacks are implemented in an array of size m. Initially, the “top” positions of the two stacks are set as: top1= -1, top2=m. In what condition that both stacks are full? • a．top2-top1=0 b．top1+1=top2 b．top1+top2=m d．top1=top2+1 • 解答：选b 0 1 …….. m-1 top1 top2
选择3 • Suppose that 5 numbers are pushed onto a stack in the order of 12345. If the output order is 35421, what is the minimum capacity of the stack should have? • a. 2 b. 3 c. 4 d. 5 • 解答：选c • 元素5是第5个push进的，是第2个pop出来的，所以深度至少为4
选择4 • When conversing the infix expression 3 *2 + 8 / 4 into its postfix , which order of stack operations will be taked? Here let P stand for “push” and O for “pop”. • a. PPPOOO b.POPOPOc.POPPOOd.PPOOPO • 解答：选c • 32*84/+
选择5 • A queue is implemented in a circular array of size 6. Now, rear and front positions are 0 and 4 respectively, afterenqueue 2 elements and dequeue 2 elements, what are the positions of rear and front respectively? • a．0 and 4 b．2 and 6 c．2 and 0 d．2 and 2 • 解答：选c • Rear:0 -> 2 front: 4 -> 0(6)
选择6 • Given a quadtree(四叉树) with 4 nodes of degree 2, 2 nodes of degree 3, 1 nodes of degree 4. The number of leaf nodes in this tree is ______. • a. 8 b. 12 c. 18 d. 20 • 解答：选b • 设叶子个数为n0，度为1的结点数为n1，度为2结点数为n2，度为4结点数为n4度为4结点数为n4 • 结点总数(记为n): n=n0+n1+n2+n3+n4 另一方面,1度,2度,3度,4度结点分别为有1-4个孩子，结点总数又可以表示为：n=nl+2n2+3n3+4n4+1故有n0+n1+n2+n3+n4=nl+2n2+3n3+4n4+1，即得n0=n2+2n3+3n4+1
选择7 • How many leaf node does a complete binary tree with 2435 nodes have? • a. 1218 b. 1217 c. 812 d. cannot be determined • 解答：选a • 深度为11的完全二叉树节点为2^11-1=2047，所以第12层的叶节点是2435-2047=388个。第11层一共有节点数2^10=1024，则该层叶节点个数是1024-388/2=830。所以叶节点一共有388+830=1218个
选择8 • In a ________, keys along each of the paths from any node to the root are ordered. • a. heap b. binary search tree c. complete tree d. none of the above • 解答：选a • B的特点是左儿子<根<右儿子，并不能保证path上节点的大小顺序
选择9 • In a min-heap with n elements, at what position may the maximum key exist? • a. n/2b. n/2 - 1c. 1d. n/2 + 2 • 解答：选d • Min-heap的特点是：1.父节点小，子节点大；2.是一个完全二叉树。所以最大值一定在最底层。 • n/2是节点n的parent，必定在倒数第二层。 • 只有d选项会出现在最底层。
选择10 • Among the following sorting algorithms, ____ has the average run time O(NlogN) with O(N) extra spaces. • a. Quick sort b. Heap sort • c. Merge sort d. Insertion sort • 解答：选c • Quicksort- O(NlogN)，O(LogN)extraspaces • Heapsort-O(NlogN)，noextraspaces • Mergesort-O(NlogN)， O(N) extraspaces • Insertionsort-O(N2)， noextraspaces
选择11 • In the following integer sequences, which is(are) NOT a heap. • a. (100,85,98,77,80,60,82,40,20,10,66) • b. (100,98,85,82,80,77,66,60,40,20,10) • c. (10,20,40,60,66,77,80,82,85,98,100) • d. (100,85,40,77,80,60,66,98,82,10,20) • 解答：选择d • D不是maxheap也不是minheap
选择12 • Among the following sorting algorithms, which has the average run time O(NlogN) with small extra space. • a. Quick sortb. Heap sortc. Merge sortd. Insertion sort • 解答：选b • Quicksort- O(NlogN)，O(LogN)extraspaces • Heapsort-O(NlogN)，noextraspaces • Mergesort-O(NlogN)， O(N) extraspaces • Insertionsort-O(N2)， noextraspaces
选择13 • Followings are the middle stages during sorting （2，12，16，88，5，10）, which sorting algorithm could the most possible used? • First Stage：2，12，16，5，10，88 • Second Stage：2，12，5，10，16，88 • Third Stage：2，5，10，12，16，88 • a．Bubbles Sort b．Selection Sort c．Heap Sort d．Insertion Sort • 解答：选a
填空 maxheap的deletemax函数 ElementTypeDeleteMax( PriorityQueue H ) { inti, Child; ElementTypeMaxElement, LastElement; MaxElement = ; H->Elements[1]//此处保存这个最大的值 LastElement = H->Elements[ H->Size-- ]; for( i = 1; i * 2 <= H->Size; i = Child ) { Child = i * 2; //找大儿子 if( Child!=H->size && H->Elements[child]<H->Elements[child+1] ) Child++; //如果右儿子大，就指向右儿子 if( LastElement < H->Elements[ Child ] ) H->Elements[i] = H->Elements[child] ;//大的向上移 else break; } H->Elements[ i ] = LastElement; return MaxElement; }
第三大题第一题 • The followings are the partial results of a binary tree’s traversals in pre-order, in-order and post-order. Please draw the corresponding tree. (13 points) • Pre-order : _B_F_ICEH_G • In-order: D_KFIA_EJC_ • Post-order: _K_FBHJ_G_A
Pre-order : 根-左-右 • In-order: 左-根-右 • Post-order: 左-右-根 • 共11个元素 A G B D F J I C H E K A B D F K I C E H J G D B K F I A H E J C G D K I F B H J E G C A
第三大题第二题 • A binary search treesix numbers 2 4 6 8 10 12 are the keys in this tree. Please: • (a)show the pre-order result of this tree. (7 points) • (b)show the resulting binary search tree after 7 is inserted.(6 points) • 解答： • Binarysearchtree 特点是左<根<右 • 所以pre-order是8,2,6,4,10,12 • 插入7 10 12 2 6 4 7 8
第四大题 • Please write a non-recursive function FirstNode() that return the first node of a binary tree in post travel order. (20 points) • structTreeNode { • ElementType Element; • TreeNode *left; • TreeNode *right; • } • TreeNode *FirstNode (TreeNode *tree) • 解答： • Post-order的遍历顺序是左-右-根 • 寻找第一个node
TreeNode *FirstNode (TreeNode *p) { while (p && (p->left || p->right)) { if (p->left) p=p->left; else p=p->right; } return p }
一些注意事项 • Project提交格式 • 打包提交. • Pdf命名规则：hqm_G18_P2.pdf • 源代码命名规则:hqm_G18_P2.c或者.cpp • 不按规则的writer会扣分 • Homework • 每次周三上机课交给我，并领取上次作业 • 至今还有同学没有交过homework • 总分计算规则 • 期末考试50分+期中考试10分+homework15分+project25分 =100分
