2020年
【2020年题09】 下列关于大根堆(至少含 2 个元素)的叙述中,正确的是( )。
I. 可以将堆视为一棵完全二叉树
II. 可以采用顺序存储方式保存堆
III. 可以将堆视为一棵二叉排序树
IV. 堆中的次大值一定在根的下一层
A. 仅 I、II B. 仅 II、III C. 仅 I、II 和 IV D. I、III 和 IV
解析:
这是一道简单的概念题。堆是一棵完全二叉树,采用一维数组存储,故 I 正确,II 正确。大根堆只要求根结点值大于左右孩子值,并不要求左右孩子值有序,III 错误。堆的定义是递归的,所以其左右子树也是大根堆,所以堆的次大值一定是其左孩子或右孩子,IV 正确。
【2020年题10】 依次将关键字 5, 6, 9, 13, 8, 2, 12, 15 插入初始为空的 4 阶 B 树后,根结点中包含的关键字是( )。
A. 8 B. 6, 9 C. 8, 13 D. 9, 12
解析:
一个 4 阶 B 树的任意非叶结点至多含有 m − 1 = 3 m-1=3 m − 1 = 3 个关键字,在关键字依次插入的过程中,会导致结点的不断分裂,插入过程如下所示。
【2020年题11】 对大部分元素已有序的数组进行排序时,直接插入排序比简单选择排序效率更高,其原因是( )。
A. 仅 I B. 仅 III C. 仅 I、II D. I、II 和 III
考虑较极端的情况,对于有序数组,直接插入排序的比较次数为 n − 1 n-1 n − 1 ,简单选择排序的比较次数始终为 1 + 2 + ⋯ + n − 1 = n ( n − 1 ) / 2 1+2+\cdots+n-1=n(n-1)/2 1 + 2 + ⋯ + n − 1 = n ( n − 1 ) /2 ,I 正确。两种排序方法的辅助空间都是 ,无差别,II 错误。初始有序时,移动次数均为 0;对于通常情况,直接插入排序每趟插入都需要依次向后挪位,而简单选择排序只需与找到的最小元素交换位置,后者的移动次数少很多,III 错误。
2021年 【2021年题10】 设数组 S[] = 892,采用最低位优先(LSD)基数排序将 S 排列成升序序列。第 1 趟分配、收集后,元素 372 之前、之后紧邻的元素分别是( )。
A. 43, 892 B. 236, 301 C. 301, 892 D. 485, 301
基数排序是一种稳定的排序方法。由于采用最低位优先(LSD)的基数排序,即第 1 趟对个位进行分配和收集的操作,因此第一趟分配和收集后的结果是 9,元素 372 之前、之后紧邻的元素分别是 301 和 892,故选 C。
【2021年题11】 将关键字 6, 9, 1, 5, 8, 4, 7 依次插入到初始为空的大根堆 H 中,得到的 H 是( )。
A. 9, 8, 7, 6, 5, 4, 1 B. 9, 8, 7, 5, 6, 1, 4
C. 9, 8, 7, 5, 6, 4, 1 D. 9, 6, 7, 5, 8, 4, 1
要熟练掌握调整堆的方法,建堆的过程如下图所示,故答案选 B。
2024年
SVG 代码库 <!-- 图1:2024年题1 链表结点结构 -->
<img src = " data:image/svg+xml;utf8,%3Csvg%20xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F2000%2Fsvg%22%20width%3D%2284%22%20height%3D%2224%22%20viewBox%3D%220%200%2084%2024%22%3E%3Crect%20x%3D%222%22%20y%3D%222%22%20width%3D%2240%22%20height%3D%2220%22%20fill%3D%22white%22%20stroke%3D%22black%22%20stroke-width%3D%221%22%2F%3E%3Crect%20x%3D%2242%22%20y%3D%222%22%20width%3D%2240%22%20height%3D%2220%22%20fill%3D%22white%22%20stroke%3D%22black%22%20stroke-width%3D%221%22%2F%3E%3Ctext%20x%3D%2222%22%20y%3D%2216%22%20font-family%3D%22serif%22%20font-size%3D%2212%22%20text-anchor%3D%22middle%22%3Edata%3C%2Ftext%3E%3Ctext%20x%3D%2262%22%20y%3D%2216%22%20font-family%3D%22serif%22%20font-size%3D%2212%22%20text-anchor%3D%22middle%22%3Enext%3C%2Ftext%3E%3C%2Fsvg%3E " />
<!-- 图2:2024年题4 邻接多重表 -->
<img src = " data:image/svg+xml;utf8,%3Csvg%20xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F2000%2Fsvg%22%20width%3D%22450%22%20height%3D%22160%22%20viewBox%3D%220%200%20450%20160%22%3E%3Cdefs%3E%3Cmarker%20id%3D%22a%22%20markerWidth%3D%226%22%20markerHeight%3D%224%22%20refX%3D%225%22%20refY%3D%222%22%20orient%3D%22auto%22%3E%3Cpolygon%20points%3D%220%200%2C%206%202%2C%200%204%22%20fill%3D%22black%22%2F%3E%3C%2Fmarker%3E%3C%2Fdefs%3E%3Cg%20font-family%3D%22serif%22%20font-size%3D%2210%22%20text-anchor%3D%22middle%22%20fill%3D%22black%22%3E%3Ctext%20x%3D%2215%22%20y%3D%2222%22%3E0%3C%2Ftext%3E%3Crect%20x%3D%2222%22%20y%3D%2212%22%20width%3D%2218%22%20height%3D%2214%22%20fill%3D%22none%22%20stroke%3D%22black%22%2F%3E%3Ctext%20x%3D%2231%22%20y%3D%2222%22%3Ea%3C%2Ftext%3E%3Crect%20x%3D%2240%22%20y%3D%2212%22%20width%3D%2218%22%20height%3D%2214%22%20fill%3D%22none%22%20stroke%3D%22black%22%2F%3E%3Ctext%20x%3D%2215%22%20y%3D%2244%22%3E1%3C%2Ftext%3E%3Crect%20x%3D%2222%22%20y%3D%2234%22%20width%3D%2218%22%20height%3D%2214%22%20fill%3D%22none%22%20stroke%3D%22black%22%2F%3E%3Ctext%20x%3D%2231%22%20y%3D%2244%22%3Eb%3C%2Ftext%3E%3Crect%20x%3D%2240%22%20y%3D%2234%22%20width%3D%2218%22%20height%3D%2214%22%20fill%3D%22none%22%20stroke%3D%22black%22%2F%3E%3Ctext%20x%3D%2215%22%20y%3D%2266%22%3E2%3C%2Ftext%3E%3Crect%20x%3D%2222%22%20y%3D%2256%22%20width%3D%2218%22%20height%3D%2214%22%20fill%3D%22none%22%20stroke%3D%22black%22%2F%3E%3Ctext%20x%3D%2231%22%20y%3D%2266%22%3Ec%3C%2Ftext%3E%3Crect%20x%3D%2240%22%20y%3D%2256%22%20width%3D%2218%22%20height%3D%2214%22%20fill%3D%22none%22%20stroke%3D%22black%22%2F%3E%3Ctext%20x%3D%2215%22%20y%3D%2288%22%3E3%3C%2Ftext%3E%3Crect%20x%3D%2222%22%20y%3D%2278%22%20width%3D%2218%22%20height%3D%2214%22%20fill%3D%22none%22%20stroke%3D%22black%22%2F%3E%3Ctext%20x%3D%2231%22%20y%3D%2288%22%3Ed%3C%2Ftext%3E%3Crect%20x%3D%2240%22%20y%3D%2278%22%20width%3D%2218%22%20height%3D%2214%22%20fill%3D%22none%22%20stroke%3D%22black%22%2F%3E%3Ctext%20x%3D%2215%22%20y%3D%22110%22%3E4%3C%2Ftext%3E%3Crect%20x%3D%2222%22%20y%3D%22100%22%20width%3D%2218%22%20height%3D%2214%22%20fill%3D%22none%22%20stroke%3D%22black%22%2F%3E%3Ctext%20x%3D%2231%22%20y%3D%22110%22%3Ee%3C%2Ftext%3E%3Crect%20x%3D%2240%22%20y%3D%22100%22%20width%3D%2218%22%20height%3D%2214%22%20fill%3D%22none%22%20stroke%3D%22black%22%2F%3E%3Cg%20stroke%3D%22black%22%20fill%3D%22none%22%3E%3Crect%20x%3D%22100%22%20y%3D%2212%22%20width%3D%2290%22%20height%3D%2214%22%2F%3E%3Cline%20x1%3D%22115%22%20y1%3D%2212%22%20x2%3D%22115%22%20y2%3D%2226%22%2F%3E%3Cline%20x1%3D%22135%22%20y1%3D%2212%22%20x2%3D%22135%22%20y2%3D%2226%22%2F%3E%3Cline%20x1%3D%22152%22%20y1%3D%2212%22%20x2%3D%22152%22%20y2%3D%2226%22%2F%3E%3Cline%20x1%3D%22170%22%20y1%3D%2212%22%20x2%3D%22170%22%20y2%3D%2226%22%2F%3E%3Crect%20x%3D%22240%22%20y%3D%2212%22%20width%3D%2290%22%20height%3D%2214%22%2F%3E%3Cline%20x1%3D%22255%22%20y1%3D%2212%22%20x2%3D%22255%22%20y2%3D%2226%22%2F%3E%3Cline%20x1%3D%22275%22%20y1%3D%2212%22%20x2%3D%22275%22%20y2%3D%2226%22%2F%3E%3Cline%20x1%3D%22292%22%20y1%3D%2212%22%20x2%3D%22292%22%20y2%3D%2226%22%2F%3E%3Cline%20x1%3D%22310%22%20y1%3D%2212%22%20x2%3D%22310%22%20y2%3D%2226%22%2F%3E%3Crect%20x%3D%22100%22%20y%3D%2234%22%20width%3D%2290%22%20height%3D%2214%22%2F%3E%3Cline%20x1%3D%22115%22%20y1%3D%2234%22%20x2%3D%22115%22%20y2%3D%2248%22%2F%3E%3Cline%20x1%3D%22135%22%20y1%3D%2234%22%20x2%3D%22135%22%20y2%3D%2248%22%2F%3E%3Cline%20x1%3D%22152%22%20y1%3D%2234%22%20x2%3D%22152%22%20y2%3D%2248%22%2F%3E%3Cline%20x1%3D%22170%22%20y1%3D%2234%22%20x2%3D%22170%22%20y2%3D%2248%22%2F%3E%3Crect%20x%3D%22100%22%20y%3D%2256%22%20width%3D%2290%22%20height%3D%2214%22%2F%3E%3Cline%20x1%3D%22115%22%20y1%3D%2256%22%20x2%3D%22115%22%20y2%3D%2270%22%2F%3E%3Cline%20x1%3D%22135%22%20y1%3D%2256%22%20x2%3D%22135%22%20y2%3D%2270%22%2F%3E%3Cline%20x1%3D%22152%22%20y1%3D%2256%22%20x2%3D%22152%22%20y2%3D%2270%22%2F%3E%3Cline%20x1%3D%22170%22%20y1%3D%2256%22%20x2%3D%22170%22%20y2%3D%2270%22%2F%3E%3Crect%20x%3D%22100%22%20y%3D%2278%22%20width%3D%2290%22%20height%3D%2214%22%2F%3E%3Cline%20x1%3D%22115%22%20y1%3D%2278%22%20x2%3D%22115%22%20y2%3D%2292%22%2F%3E%3Cline%20x1%3D%22135%22%20y1%3D%2278%22%20x2%3D%22135%22%20y2%3D%2292%22%2F%3E%3Cline%20x1%3D%22152%22%20y1%3D%2278%22%20x2%3D%22152%22%20y2%3D%2292%22%2F%3E%3Cline%20x1%3D%22170%22%20y1%3D%2278%22%20x2%3D%22170%22%20y2%3D%2292%22%2F%3E%3Crect%20x%3D%22100%22%20y%3D%22100%22%20width%3D%2290%22%20height%3D%2214%22%2F%3E%3Cline%20x1%3D%22115%22%20y1%3D%22100%22%20x2%3D%22115%22%20y2%3D%22114%22%2F%3E%3Cline%20x1%3D%22135%22%20y1%3D%22100%22%20x2%3D%22135%22%20y2%3D%22114%22%2F%3E%3Cline%20x1%3D%22152%22%20y1%3D%22100%22%20x2%3D%22152%22%20y2%3D%22114%22%2F%3E%3Cline%20x1%3D%22170%22%20y1%3D%22100%22%20x2%3D%22170%22%20y2%3D%22114%22%2F%3E%3Crect%20x%3D%22240%22%20y%3D%22100%22%20width%3D%2290%22%20height%3D%2214%22%2F%3E%3Cline%20x1%3D%22255%22%20y1%3D%22100%22%20x2%3D%22255%22%20y2%3D%22114%22%2F%3E%3Cline%20x1%3D%22275%22%20y1%3D%22100%22%20x2%3D%22275%22%20y2%3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【2024年题1】 已知带头结点的非空单链表 L 的头指针为 h,结点结构为
,其中 next 是指向直接后继结点的指针。现有指针 p 和 q,若 p 指向 L 中非首且非尾的任意一个结点,则执行语句序列“q=p->next; p->next=q->next; q->next=h->next; h->next=q;”的结果是( )。
语句 q=p->next:指针 q 指向 p 的下一个结点;
语句 p->next=q->next:p 的 next 域指向 q 的下一个结点,即将 q 所指结点从链表中取出;
语句 q->next=L->next:q 的 next 域指向第一个结点(头结点的后一个结点);
语句 L->next=q:头指针指向 q 所指结点。
因此,这段代码的功能是将 q 结点移动到表头。故本题选 D。
【2024年题2】 与表达式 x+y*(z-u)/v 等价的后缀表达式是( )。
A. xyzu-v/+ B. xyzu-v/ + C. +x/y-zuv D. +x y/-zuv
(1)根据运算的先后顺序为中缀表达式全部加上括号变为(x+((y*(z-u))/v));
(2)把运算符移到对应的括号外面,变为:(x(((y(zu)-)*v)/)+;
(3)将括号全部去除后即为后缀表达式:xyzu-*v/+。故本题选 A。
【2024年题3】 若 p 、q 和 v 均为二叉树 T 中的结点,v 有两个孩子结点,T 的中序遍历序列形如:“…,p,v,q,…”,则下列叙述中,正确的是( )。
中序遍历的顺序为左子树、根结点、右子树。本题中 v 有两个孩子结点,且中序遍历中有片段 p,v,q,则 p、q 分别是 v 左、右子树中的结点。在 v 的左子树中,p 是左子树的中序遍历序列最后一个结点,则 p 是最右边的结点,故 p 没有右孩子;在 v 的右子树中,q 是右子树的中序遍历序列第一个结点,则 q 是最左边的结点,故 q 没有左孩子。故本题选 A。
【2024年题4】 若无向图 G = (V, E)的邻接多重表如下图所示,则 G 中顶点 b 与 d 的度分别是( )。
A. 0,2 B. 2,4 C. 2,5 D. 3,4
本题有两种做法,一种是根据邻接多重表还原图,另一种是直接数出 abcd 对应的 1234 出现的次数,由下图可知,b 对应的 1 出现了 2 次,度为 2,d 对应的 3 出现了 4 次,度为 4。故本题选 B。
【2024年题5】 下列数据结构中,不适合直接使用折半查找的是( )。
I.有序链表 II.无序数组 III.有序静态链表 IV.无序静态链表
A.仅 I 、III B.仅 II、IV C.仅 II、III、IV D.I、II、III、IV
折半查找,也称二分查找,在某些情况下相比于顺序查找,使用折半查找算法的效率更高。但是该算法的使用的前提是静态查找表中的数据必须是有序的。也就是说,在使用折半查找算法查找数据之前,应该首先把该表的数据按照所查的关键字进行排序。链表不支持随机查找,故 I 不适用,无序数组不适用折半查找,故 II 不适用。静态链表是用数组来实现链式存储结构,目的是方便在不设指针类型的高级程序设计语言中使用链式结构,故 III、IV 不适用。故本题选 D。
【2024年题6】 KMP 算法使用修正后的 next 数组进行模式匹配,模式串 S= “aabaab”,当主串中某字符与 S 中某字符失配时,S 将向右滑动的最长距离是( )。
本题考查 KMP 算法中 nextval 的应用。
传统 KMP 算法使用 next 数组,会存在 j=next[j]的情况,因此采用 next_val 来优化,处理结果如下:
当主串中某个字符与 S 中某字符失配时,有 6 种情况,分别为下标 1-6 不匹配:
【2024年题7】 一棵二叉搜索树如题 7 图所示,k1、k2、k3 分别是对应结点保存的关键字,子树 T 的任一结点中保存的关键字 x 满足的是( )。
A. x < k1 B. x > k2 C. k1 < x < k3 D. k3 < x < k2
本题考查二叉搜索树的基本性质。二叉搜索树,即二叉排序树。由二叉排序(搜索)树的性质:若它的左子树不空,则左子树上所有结点的值均小于它的根结点的值;若它的右子树不空,则右子树上所有结点的值均大于它的根结点的值。那么大小顺序应该是:K1\
进行两次删除操作后,得到的新堆是 20,19,15,5,8,12;故本题选 B。
【2024年题10】 现有由关键字组成的 3 个有序序列(3,5)、(7,9)和(6),若按从左至右的次序选择有序序列进行二路归并排序,则关键字之间的总比较次数是( )。
二路归并排序的思想是假设初始序列含有 n 个记录,则可看成是 n 个有序的子序列,每个子序列的长度为 1,然后两两归并,得到个长度为 2 或 1 的有序子序列;再两两归并,……,如此重复,直至得到一个长度为 n 的有序序列为止。
3 和 7 进行比较,3 存入序列得(3);(1 次比较)
5 和 7 进行比较,5 存入序列得(3,5);(1 次比较)
子序列(3,5)处理完毕,将剩下的依次存入得(3,5,7,9)。
3 和 6 进行比较,3 存入序列得(3);(1 次比较)
5 和 6 进行比较,5 存入序列得(3,5);(1 次比较)
7 和 6 进行比较,6 存入序列得(3,5,6);(1 次比较)
子序列(6)处理完毕,将剩下的依次存入得(3,5,6,7,9)。
【2024年题11】 在外排序中,利用败者树对初始为升序的归并段进行多路归并,败者树中记录“冠军”的结点保存的是( )。
A.最大关键字 B.最小关键字 C.最大关键字所在的归并段号 D.最小关键字所在的归并段号
败者树是胜者树的一种变体。在胜者树中,中间结点记录的是胜者的标号,而在败者树中,中间结点记录的是败者的标号。败者树与胜者树很类似,通过比赛使胜者晋级,直到决出冠军。但是败者树的非叶子节点记录的是败者(最小关键字)的归并段号,且需要增加一个节点记录比赛的胜者。故本题选 D。
心得笔记
建堆的过程:教材讲什么稀奇古怪的筛选,让人完全无法理解。实际上是把不合格的结点向下沉。从最后一个非叶子结点开始,每个选中的叶子结点只向下比较,向下交换。然后再比较下一个位置,也就是上一个非叶子结点。但每一趟调整都必须一路向下比较,如果不对就换。
堆排序的时间复杂度,求和式子是怎么算的
24年题,按从左至右的次序选择有序序列,这个给我干懵了,其实不就是有序序列按从左到右归并吗?来个选择让人搞不懂了。。。
O ( 1 )
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" />
Dors Dors 是花野猫开发为知识工作者打造的数字花园应用,包含的博客、个人记事本、及其他实用功能。