Questions
XLMC0202501 Topic 5 Quiz
Multiple choice
Which of the following types of list can be implemented so that algorithms for deletion of the element at the beginning and the end of the list are both O(1) time? (check all that apply)
Options
A.Doubly-linked list
B.Singly linked list
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Step-by-Step Analysis
When evaluating data structures for constant-time deletions at both ends, we need to consider how pointers are arranged and what information is stored.
Option 1: Doubly-linked list. In a doubly-linked list, each node has pointers to both its successor and its predecessor. With direct references to the head and to the tail, removing the first elem......Login to view full explanationLog in for full answers
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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:
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
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