import Foundation
let pattern = #"(?=^[^1]*(?:1,1,1,1,1|1(?:(?:\D+[^1]){9}\D+1){4})[^1]*$)(?=^[^2]*(?:2,2,2,2|2(?:(?:\D+[^2]){9}\D+2){3})[^2]*$)(?=^[^3]*(?:3,3,3|3(?:(?:\D+[^3]){9}\D+3){2})[^3]*$)(?=^[^4]*(?:4,4,4|4(?:(?:\D+[^4]){9}\D+4){2})[^4]*$)(?=^[^5]*(?:5,5|5(?:\D+[^5]){9}\D+5)[^5]*$)(?:(?:^\[|)\[[0-5](?:,[0-5]){9}\](?:,|\]$)){10}"#
let regex = try! NSRegularExpression(pattern: pattern)
let testString = #"[[0,1,1,1,1,1,0,0,0,0],[0,0,0,0,0,0,0,2,0,0],[0,0,0,0,0,0,0,2,0,0],[0,0,0,0,0,0,0,2,0,0],[0,3,3,3,0,0,0,2,0,0],[0,0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0,0],[0,0,5,0,0,0,4,4,4,0],[0,0,5,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0,0]]"#
let stringRange = NSRange(location: 0, length: testString.utf16.count)
let matches = regex.matches(in: testString, range: stringRange)
var result: [[String]] = []
for match in matches {
var groups: [String] = []
for rangeIndex in 1 ..< match.numberOfRanges {
let nsRange = match.range(at: rangeIndex)
guard !NSEqualRanges(nsRange, NSMakeRange(NSNotFound, 0)) else { continue }
let string = (testString as NSString).substring(with: nsRange)
groups.append(string)
}
if !groups.isEmpty {
result.append(groups)
}
}
print(result)
Please keep in mind that these code samples are automatically generated and are not guaranteed to work. If you find any syntax errors, feel free to submit a bug report. For a full regex reference for Swift 5.2, please visit: https://developer.apple.com/documentation/foundation/nsregularexpression