Word break problem: Difference between revisions
→{{header|Wren}}
Line 1,727:
abcdd:
did not parse with given words</pre>
=={{header|Visual Basic .NET}}==
{{trans|Java}}
<lang vbnet>Module Module1
Structure Node
Private ReadOnly m_val As String
Private ReadOnly m_parsed As List(Of String)
Sub New(initial As String)
m_val = initial
m_parsed = New List(Of String)
End Sub
Sub New(s As String, p As List(Of String))
m_val = s
m_parsed = p
End Sub
Public Function Value() As String
Return m_val
End Function
Public Function Parsed() As List(Of String)
Return m_parsed
End Function
End Structure
Function WordBreak(s As String, dictionary As List(Of String)) As List(Of List(Of String))
Dim matches As New List(Of List(Of String))
Dim q As New Queue(Of Node)
q.Enqueue(New Node(s))
While q.Count > 0
Dim node = q.Dequeue()
REM check if fully parsed
If node.Value.Length = 0 Then
matches.Add(node.Parsed)
Else
For Each word In dictionary
REM check for match
If node.Value.StartsWith(word) Then
Dim valNew = node.Value.Substring(word.Length, node.Value.Length - word.Length)
Dim parsedNew As New List(Of String)
parsedNew.AddRange(node.Parsed)
parsedNew.Add(word)
q.Enqueue(New Node(valNew, parsedNew))
End If
Next
End If
End While
Return matches
End Function
Sub Main()
Dim dict As New List(Of String) From {"a", "aa", "b", "ab", "aab"}
For Each testString In {"aab", "aa b"}
Dim matches = WordBreak(testString, dict)
Console.WriteLine("String = {0}, Dictionary = {1}. Solutions = {2}", testString, dict, matches.Count)
For Each match In matches
Console.WriteLine(" Word Break = [{0}]", String.Join(", ", match))
Next
Console.WriteLine()
Next
dict = New List(Of String) From {"abc", "a", "ac", "b", "c", "cb", "d"}
For Each testString In {"abcd", "abbc", "abcbcd", "acdbc", "abcdd"}
Dim matches = WordBreak(testString, dict)
Console.WriteLine("String = {0}, Dictionary = {1}. Solutions = {2}", testString, dict, matches.Count)
For Each match In matches
Console.WriteLine(" Word Break = [{0}]", String.Join(", ", match))
Next
Console.WriteLine()
Next
End Sub
End Module</lang>
{{out}}
<pre>String = aab, Dictionary = System.Collections.Generic.List`1[System.String]. Solutions = 4
Word Break = [aab]
Word Break = [a, ab]
Word Break = [aa, b]
Word Break = [a, a, b]
String = aa b, Dictionary = System.Collections.Generic.List`1[System.String]. Solutions = 0
String = abcd, Dictionary = System.Collections.Generic.List`1[System.String]. Solutions = 2
Word Break = [abc, d]
Word Break = [a, b, c, d]
String = abbc, Dictionary = System.Collections.Generic.List`1[System.String]. Solutions = 1
Word Break = [a, b, b, c]
String = abcbcd, Dictionary = System.Collections.Generic.List`1[System.String]. 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 = System.Collections.Generic.List`1[System.String]. Solutions = 2
Word Break = [ac, d, b, c]
Word Break = [a, c, d, b, c]
String = abcdd, Dictionary = System.Collections.Generic.List`1[System.String]. Solutions = 2
Word Break = [abc, d, d]
Word Break = [a, b, c, d, d]</pre>
=={{header|Wren}}==
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