Word break problem: Difference between revisions
Added 11l
MaiconSoft (talk | contribs) (Added Delphi example) |
Alextretyak (talk | contribs) (Added 11l) |
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{{Template:Strings}}
<br><br>
=={{header|11l}}==
{{trans|D}}
<lang 11l>T Node
String val
[String] parsed
F (val, [String] &parsed = [String]())
.val = val
.parsed = parsed
F word_break(s, dictionary)
[[String]] matches
V queue = Deque([Node(s)])
L !queue.empty
V node = queue.pop_left()
I node.val.empty
matches [+]= node.parsed
E
L(word) dictionary
I node.val.starts_with(word)
V val_new = node.val[word.len .< node.val.len]
V parsed_new = copy(node.parsed)
parsed_new [+]= word
queue [+]= Node(val_new, &parsed_new)
R matches
F process(d, test_strings)
L(test_string) test_strings
V matches = word_break(test_string, d)
print((‘String = ’test_string‘, Dictionary = ’String(d)‘. Solutions =’)‘ ’matches.len)
L(match) matches
print(‘ Word Break = ’match)
print()
V d = [‘a’, ‘aa’, ‘b’, ‘ab’, ‘aab’]
process(d, [‘aab’, ‘aa b’])
d = [‘abc’, ‘a’, ‘ac’, ‘b’, ‘c’, ‘cb’, ‘d’]
process(d, [‘abcd’, ‘abbc’, ‘abcbcd’, ‘acdbc’, ‘abcdd’])</lang>
{{out}}
<pre style="height:45ex;overflow:scroll">
String = aab, Dictionary = [a, aa, b, ab, aab]. Solutions = 4
Word Break = [aab]
Word Break = [a, ab]
Word Break = [aa, b]
Word Break = [a, a, b]
String = aa b, Dictionary = [a, aa, b, ab, aab]. Solutions = 0
String = abcd, Dictionary = [abc, a, ac, b, c, cb, d]. Solutions = 2
Word Break = [abc, d]
Word Break = [a, b, c, d]
String = abbc, Dictionary = [abc, a, ac, b, c, cb, d]. Solutions = 1
Word Break = [a, b, b, c]
String = abcbcd, Dictionary = [abc, a, ac, b, c, cb, d]. Solutions = 3
Word Break = [abc, b, c, d]
Word Break = [a, b, cb, c, d]
Word Break = [a, b, c, b, c, d]
String = acdbc, Dictionary = [abc, a, ac, b, c, cb, d]. Solutions = 2
Word Break = [ac, d, b, c]
Word Break = [a, c, d, b, c]
String = abcdd, Dictionary = [abc, a, ac, b, c, cb, d]. Solutions = 2
Word Break = [abc, d, d]
Word Break = [a, b, c, d, d]
</pre>
=={{header|Aime}}==
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