import Foundation
let pattern = #"((\d*[1-9]\d*)[0](\d*[1-9]\d*)|(\d*[2-9]\d*)[1](\d*[2-9]\d*)|(\d*[3-9]\d*)[2](\d*[3-9]\d*)|(\d*[4-9]\d*)[3](\d*[4-9]\d*)|(\d*[5-9]\d*)[4](\d*[5-9]\d*)|(\d*[6-9]\d*)[5](\d*[6-9]\d*)|(\d*[7-9]\d*)[6](\d*[7-9]\d*)|(\d*[8-9]\d*)[7](\d*[8-9]\d*)|(\d*[9]\d*)[8](\d*[9]\d*))|((\d*[0-8]\d*)[9](\d*[0-8]\d*)|(\d*[0-7]\d*)[8](\d*[0-7]\d*)|(\d*[0-6]\d*)[7](\d*[0-6]\d*)|(\d*[0-5]\d*)[6](\d*[0-5]\d*)|(\d*[0-4]\d*)[5](\d*[0-4]\d*)|(\d*[0-3]\d*)[4](\d*[0-3]\d*)|(\d*[0-2]\d*)[3](\d*[0-2]\d*)|(\d*[0-1]\d*)[2](\d*[0-1]\d*)|(\d*[0]\d*)[1](\d*[0]\d*))"#
let regex = try! NSRegularExpression(pattern: pattern)
let testString = #"""
123
234567
0123456789
87654
321
985
346
320
12342
7655338
9876789
2000000001
"""#
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