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Word break problem: Difference between revisions
Add Swift
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(Add Swift) |
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mississippi: ["miss", "is", "sip", "pi"]
</pre>
=={{header|Swift}}==
{{trans|Rust}}
<lang swift>infix operator ??= : AssignmentPrecedence
@inlinable
public func ??= <T>(lhs: inout T?, rhs: T?) {
lhs = lhs ?? rhs
}
private func createString(_ from: String, _ v: [Int?]) -> String {
var idx = from.count
var sliceVec = [Substring]()
while let prev = v[idx] {
let s = from.index(from.startIndex, offsetBy: prev)
let e = from.index(from.startIndex, offsetBy: idx)
sliceVec.append(from[s..<e])
idx = prev
}
return sliceVec.reversed().joined(separator: " ")
}
public func wordBreak(str: String, dict: Set<String>) -> String? {
let size = str.count + 1
var possible = [Int?](repeating: nil, count: size)
func checkWord(i: Int, j: Int) -> Int? {
let s = str.index(str.startIndex, offsetBy: i)
let e = str.index(str.startIndex, offsetBy: j)
return dict.contains(String(str[s..<e])) ? i : nil
}
for i in 1..<size {
possible[i] ??= checkWord(i: 0, j: i)
guard possible[i] != nil else {
continue
}
for j in i+1..<size {
possible[j] ??= checkWord(i: i, j: j)
}
if possible[str.count] != nil {
return createString(str, possible)
}
}
return nil
}
let words = [
"a",
"bc",
"abc",
"cd",
"b"
] as Set
let testCases = [
"abcd",
"abbc",
"abcbcd",
"acdbc",
"abcdd"
]
for test in testCases {
print("\(test):")
print(" \(wordBreak(str: test, dict: words) ?? "did not parse with given words")")
}</lang>
{{out}}
<pre>abcd:
a b cd
abbc:
a b bc
abcbcd:
a bc b cd
acdbc:
a cd bc
abcdd:
did not parse with given words</pre>
=={{header|zkl}}==
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