Data Structures and Algorithms in Java[1].pdf - Fulvio Frisone

Figure 9.7: Realization of a dictionary D by means
of an ordered search table. We show only the keys for
this dictionary, so as to highlight their ordering.
The space requirement of an ordered search table is O(n), which is similar to the
list-based dictionary implementation (Section 9.3.1), assuming we grow and shrink
the array supporting the array list S to keep the size of this array proportional to the
number of entries in S. Unlike an unordered list, however, performing updates in a
search table takes a considerable amount of time. In particular, performing the
insert(k,v) operation in a search table requires O(n) time, since we need to shift
up all the entries in the array list with key greater than k to make room for the new
entry (k, v). A similar observation applies to the operation remove (k), since it
takes O(n) time to shift all the entries in the array list with key greater than k to
close the "hole" left by the removed entry (or entries). The search table
implementation is therefore inferior to the log file in terms of the worst-case
running times of the dictionary update operations. Nevertheless, we can perform the
find method much faster in a search table.
Binary Search
A significant advantage of using an ordered array list S to implement a dictionary
D with n entries is that accessing an element of S by its index takes O(1) time. We
recall, from Section 6.1, that the index of an element in an array list is the number
of elements preceding it. Thus, the first element in S has index 0, and the last
element has index n − 1.
The elements stored in S are the entries of dictionary D, and since S is ordered, the
entry at index i has a key no smaller than the keys of the entries at indices 0,…, i
− 1, and no larger than the keys of the entries at indices i + 1,…, n − 1. This
observation allows us to quickly "home in" on a search key k using a variant of
the children's game "high-low." We call an entry of D a candidate if, at the
current stage of the search, we cannot rule out that this entry has key equal to k.
The algorithm maintains two parameters, low and high, such that all the
candidate entries have index at least low and at most high in S. Initially, low =
0 and high = n − 1. We then compare k to the key of the median candidate e, that
is, the entry e with index
mid = (low + high)/2.
We consider three cases:
553