12/15/06
20:02:28 1
practice.sol
CS 61B: Selected Solutions (Practice for the Final Exam)
[1] [a] insertion sort
71808294
17808294
01788294
01278894
01247889
[b] selection sort
71808294
01878294
01278894
01248897
01247898
01247889
[c] mergesort
71808294
17,80,8294
17,08,8294
17,08,28,94
17,08,28,49
0178,28,49
0178,2489
01247889
[d] quicksort
71808294
21808794
21088794
210|4|8798
0|12|4|8798
012|4|7898
012|4|7|8|98
01247889
[e] heapsort
71808294 \
81807294 |
82807194 | bottomUpHeap()
82897104 |
82897140 /
08897241
01897842
01298874
01249887
01247898
01247898
01247889
[2] Extend each item so that it has a "secondary key," which is the index of
the item in the initial array/list. If two items have the same primary
key, the tie is broken using the secondary key, so no two items are ever
considered equal.
5 8 7 5 8 8 3 7 => 5/0 8/1 7/2 5/3 8/4 8/5 3/6 7/7
[3] [a] DFS: abdcegfhi
BFS: abcdegifh
[b] DFS: efihgcdba
[c] gh, ac, hi, ab, cd, cg, fg, ce (cg may come before cd instead)
[4] [a] [i] O(|V|^2) [ii] O(|V|)
[b] Visited in preorder. Finished in postorder.
[c] With BFS, it’s done exactly the same as with DFS.
(See Homework 9 for a description of how it’s done with the latter.)
[d] for (each vertex v in the graph) {
if (v has not been visited) {
increment the count of connected components
perform DFS on v, thereby marking all vertices in its
connected component as having been visited
}
}
This algorithm requires that you don’t "unmark" the marked vertices
between calls to DFS.
[5] [a] 3 12
/ \ /
1 12 3
/\ / / \
0 2 11 1 11
/ /\ /
=> 5 => 0 2 5
/ \ / \
4 7 4 7
/ \ / \
6 9 6 9
/ \ / \
8 10 8 10
[b] 4.5
/ \
3 5
/ \ \
1 4 11
/ \ / \
0 2 7 12
/ \
6 9
/ \
8 10
