import Foundation
let pattern = #"((((([1-9][0-9]((0[48])|([13579][26])|([2468][048])))|(([1-9]((0[48])|([13579][26])|([2468][048]))))|(([13579][26])|([2468][048]))|([48]))|(((([13579][26])|([2468][048]))|[48])00))\/2\/(([1-9])|(1[0-9])|([2][0-9])))|(([1-9][0-9]{0,3})\/((([13578]|1[02])\/([1-9]|[12][0-9]|3[01]))|((([469]|11)\/([1-9]|([12][0-9])|(30)))|(2\/(([1-9])|(1[0-9])|([2][0-8]))))))) ((([02468])|([1][02468])|(2[02])):00)"#
let regex = try! NSRegularExpression(pattern: pattern, options: .anchorsMatchLines)
let testString = #"""
判断时间的合法性
PTA作业链接
https://pintia.cn/problem-sets/1244999401232809984/problems/1245332988779970560
题目要求链接
https://images.ptausercontent.com/218e185f-05fe-4a17-be9b-640b21b74373.pdf
2015/8/5 2:00
"""#
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