如何在Swift 2中搜索String

时间:2016-05-04 21:15:53

标签: swift swift2

我是swift的新手。希望解决这个难题......任何人都可以告诉我如何使用一个字符串的旧式indexOf参数传递给另一个字符串?

Can anybody solve this?

2 个答案:

答案 0 :(得分:1)

您需要检查字符串的字符。像这样:

str.characters.indexOf("a")

这将返回角色的索引" a"在字符串中。

答案 1 :(得分:0)

第1部分:

hash功能:

func hash(s: String) -> Int {
  let letters = "acdegilmnoprstuw"
  return (s.characters.startIndex..<s.characters.endIndex).reduce(7) { (h, currentIndex) -> Int in
    let codeUnit = s.characters[currentIndex]
    let startIndex = letters.characters.startIndex
    let endIndex = letters.characters.indexOf(codeUnit)!
    let index = startIndex.distanceTo(endIndex)
    return h * 37 + index
  }
}

let sevenLetter = hash("leepadg")
print(sevenLetter) // prints 680131659347

第2部分:

// based on https://blog.hjm.im/hashing-puzzle/
func reverseHash(hash: Int, allowedCharacters: [Character]) -> String {
  var hash = hash
  var indices = [Int: Int]()
  var i = 0
  while hash > 37 {
    indices[i] = hash % 37
    hash = hash / 37
    i += 1
  }
  return (indices.count - 1).stride(through: 0, by: -1).reduce("") {
    $0 + String(allowedCharacters[indices[$1]!])
  }
}

let chars: [Character] = ["a", "c", "d", "e", "g", "i", "l", "m", "n", "o", "p", "r", "s", "t", "u", "w"]
print(reverseHash(680131659347, allowedCharacters: chars)) // prints "leepadg"

我将它作为练习留给读者用正确的哈希替换哈希:)