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《数据结构》试题一、选择题(每小题2分,共30分)1. 若某线性表中最常用的操作是取第i 个元素和找第i个元素的前趋元素,则采用( )存储方式最节省时间。A、单链表 B、双链表 C、单向循环 D、顺序表2. 串是任意有限个( )A、符号构成的序列 B、符号构成的集合C、字符构成的序列 D、字符构成的集合3. 设矩阵A(aij ,l≤i,j≤ 10)的元素满足:aij≠0(i≥j, l≤i, j≤ 10)aij=0 (i。( )i6.对有向图G,如果从任一顶点出发进行一次深度优先或广度优先搜索就能访问每个顶点,则该图一定是完全图。( )7.“顺序查找法”是指在顺序表上进行查找的方法。( )8.向二叉排序树插入一个新结点时,新结点一定成为二叉排序树的一个叶子结点。()9.键值序列{A,C,D,E,F,E,F}是一个堆。10.二路归并时,被归并的两个子序列中的关键字个数一定要相等。() 三、填空题(每小题2分,共20分)1.在带有头结点的单链表L中,若要删除第一个结点,则需执行下列三条语句:________;L->next=U->next;free(U);2.有一个长度为20的有序表采用二分查找方法进行查找,共有______个元素的查找长度为3。3.采用冒泡排序对有n个记录的表A按键值递增排序,若L的初始状态是按键值递增,则排序过程中记录的比较次数为_____。若A的初始状态为递减排列,则记录的交换次数为_______。4.在无头结点的双链表中,指针P所指结点是第一个结点的条件是______。5.G为无向图,如果从G的某个顶点出发,进行一次广度优先搜索,即可访问图的每个顶点,则该图一定是_____图。6.如果一个有向图中没有______,则该图的全部顶点可能排成一个拓扑序列。7.深度为8(根的层次号为1)的满二叉树有______个叶子结点。 8.将一棵有100个结点的完全二叉树按层编号,则编号为49的结点X,其双亲PARENT(X)的编号为_______。9.设某闭散列表HT未满,散列函数H(KEY)为键值第一字母在字母表中的序号,处理冲突方法为线性探测法,请在下列算法划线处填上适当内容,以实现按键值第一字母的顺序输出闭散列表中所有键值的算法。void printword(keytype HT[m]) { for(i=1;i<=26;i++) { j=i; while(____________________) { if (____________________) printf(“datatype”,HT[j]); j=(j+1)% m; } } }10.设有一个链队,结点结构为data|next,front为队头指针,rear为队尾指针,当执行入队操作时需执行下列语句:malloc(p);p->data=x; p->next=NULL;________________;________________; 四、简答题:(每小题4分,共20分)1. 对于一个有10000个结点的二叉树,树叶最多有多少个?最少有多少个?2. 已知一棵二叉树的中序序列和后序序列分别为: DBGEACHF和DGEBHFCA,则该二叉树的前序序列是什么?3. 设有1000个无序的元素,需排出前10个最大(小)的元素,你认为采用哪种排序方法最快?为什么?4. 在KMP算法中,已知模式串为ADABCADADA ,请写出模式串的next[j]函数值。5. 中序遍历的递归算法平均空间复杂度为多少? 五、 算法设计题(每小题10分,共20分)1. 试编写一个算法,判断一给定的整型数组a[n]是不是一个堆。2. 一棵二叉树的繁茂度定义为各层结点数的最大值与树的高度的乘积。试写一高效算法,求二叉树的繁茂度。参考答案一、选择题1、D 2、C 3、D 4、C 5、D 6、D 7、A 8、B 9、D 10、C 11、D 12、C 13、B 14、C 15、D 二、判断题 1. √ 2. × 3. √ 4. √ 5. × 6. × 7. × 8. √ 9. √ 10. × 三、填空题1.U=L - > next2.4。 3.n-1、n(n-1)/2。4.p - > prior = NULL。5.连通6.回路或环7.28-1 = 27 = 1288.249.HT[j]!=NULL或HT[j]不为空、H(HT[j])=I10.rear - > next = p、rear = p四、简答题:1. 答: 最多是完全二叉树的形态,即5000个叶子;最少是单支树的形态,即1个叶子。2.答:是:ABDEGCFH3. 答:用锦标赛排序或堆排序很合适,因为不必等全部元素排完就能得到所需结果,时间效率为O(nlog2n); 即O(1000log21000)=O(10000) 锦标赛排序的准确比较次数为:n-1+9log2n=999+9log21000=999+9×10=1089堆排序的准确比较次数为:n-1+9log2n=999+9log21000=999+9×10=1089若用冒泡排序也较快,最多耗费比较次数为(n-1+n-2+……+n-10)=10n-55=10000-55=9945(次)4. 答: 01121123435. 答: 要考虑递归时占用了栈空间,但递归次数最多不超过树的高度,所以空间复杂度为O(log2n) 五、 算法设计题1.解:提示:堆的定义是:kia[2*i]|| a[i]>a[2*i+1])return(-1);return(minleap)};else { for( i=1; i<=n/2; i++) if (a[i]lchild) EnQueue(Q,{r.node->lchild, r.layer+1}); if(r.node->rchild) EnQueue(Q,{r.node->rchild, r.layer+1}); } //按层序入队时要随时标注结点所在层号 h=r.layer; //最后一个队列元素所在层就是树的高度 for(maxn=count[0], i=1; h; i++) if(count[i]>maxn) maxn=count[i]; //求出哪一层结点数最多 return (h*maxn)} // Width
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