题目
题目

COMP10002_2025_SM2 Practice Exam

匹配题

The following declarations are exactly the same as the ones used in the file lists.h that was discussed in class. typedef struct node node_t; struct node { data_t data; node_t *next; }; typedef struct { node_t *head; node_t *foot; } list_t; A student wrote the following function to reverse the nodes in a list by reassigning all the pointers, so that the node (and data item) that used to be at the head of the list is now at the foot, and the node (and data item) that used to be at the foot of the list is now at the head. But four of the student's assignment statements have been lost. list_t *reverse(list_t *list) { node_t *curr, *prev, *next; assert(list); prev = NULL; curr = list->head; while (curr) { // line A // line B // line C curr = next; } list->foot = list->head; // line D return list; } Match the line locations on the left with the correct assignment statements on the right. 1: line A 2: line B 3: line C 4: line D

选项
A.list->head = list->foot
B.list->head = prev
C.list->head = list->head->next
D.curr->next = prev
E.curr->next = curr
F.curr = prev
G.next = next->next
H.prev = curr
I.curr = next
J.next = curr->next
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标准答案
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思路分析
We’re given a classic in-place reversal of a singly linked list and asked to match each line position (A, B, C, D) with the appropriate assignment statement. Line A: next = curr->next - This step saves the next node before we overwrite curr's next pointer. In a reversal loop, we typically need to remember where to continue from, so we store curr->next into a temporary variable named next. This is essential to avoid losing the rest of the list as we redirect pointers. - The other options here would either move pointers prematurely or reference the wrong source; for example......Login to view full explanation

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类似问题

The following declarations are exactly the same as the ones used in the file lists.h that was discussed in class. typedef struct node node_t; struct node { data_t data; node_t *next; }; typedef struct { node_t *head; node_t *foot; } list_t; A student wrote the following function to remove every second item from a list, keeping the first, third, fifth (and so on) original items, and deleting and freeing the second, fourth, sixth (and so on) original items. But four of the student's assignment statements have been lost. list_t *drop_half(list_t *list) { node_t *curr, *fllw; // line A while (curr && curr->next) { list->foot = curr; // line B // line C free(fllw); fllw = NULL; // line D } if (curr) { list->foot = curr; } return list; } Match the line locations on the left with the correct assignment statements on the right. 1: Line A contains: 2: Line B contains: 3: Line C contains: 4: Line D contains:

In a doubly linked list, each node contains:

The following declarations are exactly the same as the ones used in the file lists.h that was discussed in class. typedef struct node node_t; struct node { data_t data; node_t *next; }; typedef struct { node_t *head; node_t *foot; } list_t; A student wrote the following function to walk through lists, swapping each adjacent pair of "even position, odd position" elements so that the first element in each pair switches to the odd position, and the second element in each pair switches to the even position. The first item in the list is counted as being position zero, which is even. None of the data items are moved, and the pairwise swaps are accomplished by pointer assignments. For example, if the original list was 10->16->15->18->17->20->19, the list returned from the function would be linked together in the order 16->10->18->15->20->17->19. In an odd-length list the last item stays in the final position. But four of the student's assignment statements have been lost, marked by four lines with comments and labels. list_t *rearrange(list_t *list) { node_t *prev, *curr, *then, *aftr; // --------------> line A curr = list->head; while (curr && curr->next) { // two more nodes exist then = curr->next; // --------------> line B // now rearrange the pointers if (prev) { prev->next = then; } else { list->head = then; } then->next = curr; curr->next = aftr // then step forwards to the following pair prev = curr; // --------------> line C } if (!curr) { // there was an even number of nodes, so // need to adjust the foot pointer too // --------------> line D } return list; } Match the line locations on the left with the correct assignment statements on the right. 1: Line A contains: 2: Line B contains: 3: Line C contains: 4: Line D contains:

In Class Activity 2 _1 What would be the state of self.tail in the RecursiveSinglyLinkedList after adding a new node to an initially empty list using the add method?

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