Levenshtein distance: Difference between revisions

Content added Content deleted
m (→‎{{header|F#}}: Corrected header as suggested on the Count examples/Full list/Tier 4 talk page)
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<lang dyalect>func levenshtein(s, t) {
<lang dyalect>func levenshtein(s, t) {
var n = s.len()
var n = s.Length()
var m = t.len()
var m = t.Length()
var d = Array.empty(n + 1, () => Array.empty(m + 1))
var d = Array.Empty(n + 1, () => Array.Empty(m + 1))

if n == 0 {
if n == 0 {
return m
return m
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d[i][0] = i
d[i][0] = i
}
}

for j in 0..m {
for j in 0..m {
d[0][j] = j
d[0][j] = j
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}
}
}
}

d[n][m]
d[n][m]
}
}

func run(x, y) {
func run(x, y) {
print("\(x) -> \(y) = \(levenshtein(x, y))")
print("\(x) -> \(y) = \(levenshtein(x, y))")
}
}

run("rosettacode", "raisethysword")</lang>
run("rosettacode", "raisethysword")</lang>