Swift 4 String 练习

练习1

• 实现一个函数，接收一个字符串，计算每个字符出现的次数。
• 按照字符出现的次数排序。
• 输出直方图。

func printCharacterCount(for string: String) {
  guard string.count > 0 else { return }

  var counts: [Character: Int] = [:]

  for i in string {
    counts[i, default: 0] += 1
  }

  let sortedKeys = counts.keys.sorted { counts[$0]! > counts[$1]! }

  let max = counts[sortedKeys.first!]!

  for key in sortedKeys {
    let value = counts[key]!
    let widthOfHashes = (value * 20) / max
    let hashes = String(repeating: "#", count: widthOfHashes)
    print("\(key) : \(hashes) \(value)")
  }
}

printCharacterCount(for: "The quick brown fox jumps over the lazy dog")

技术点分析：

• guard语法糖，排除错误的情况。
• counts[i, default: 0] += 1，default关键字实现默认值。
• "!"强制解包。
• String(repeating: "#", count: widthOfHashes)初始化字符串。

输出：

  : #################### 8
o : ########## 4
e : ####### 3
u : ##### 2
h : ##### 2
r : ##### 2
w : ## 1
b : ## 1
c : ## 1
q : ## 1
y : ## 1
v : ## 1
j : ## 1
p : ## 1
x : ## 1
f : ## 1
k : ## 1
t : ## 1
z : ## 1
d : ## 1
a : ## 1
i : ## 1
T : ## 1
m : ## 1
s : ## 1
n : ## 1
l : ## 1
g : ## 1

练习2

• 计算字符串中的单词数量

func numberOfWords(in sentence: String) -> Int {
  var count = 0
  var inWord = false

  for character in sentence {
    if character == " " {
      if inWord {
        count += 1
      }
      inWord = false
    } else {
      inWord = true
    }
  }

  // We need to add 1 to count the final word if there was at least 1 character
  if inWord {
    count += 1
  }

  return count
}

let wordCount = numberOfWords(in: "  The  quick brown fox jumps over the lazy dog")
print(wordCount)

练习3

• 格式化字符串
• 输入："Galloway, Matt"，输出："Matt Galloway"。

func sanitized(name: String) -> String? {
  guard let indexOfComma = name.index(of: ",") else {
    return nil
  }

  let indexAfterCommaSpace = name.index(indexOfComma, offsetBy: 2)

  let lastNameSubstring = name[..<indexOfComma]
  let firstNameSubstring = name[indexAfterCommaSpace...]

  return firstNameSubstring + " " + lastNameSubstring
}

if let sanitizedName = sanitized(name: "Galloway, Matt") {
  print(sanitizedName)
}

技术点：

• 字符串索引
• 子串，范围区间的使用

练习4

• 实现components(separatedBy:) 函数。

func splitting(_ string: String, delimiter: Character) -> [String] {
  var returnArray: [String] = []
  var lastWordIndex = string.startIndex

  for i in string.indices {
    if string[i] == delimiter {
      let substring = string[lastWordIndex..<i]
      returnArray.append(String(substring))
      lastWordIndex = string.index(after: i)
    }
  }

  // Add the final word
  let substring = string[lastWordIndex..<string.endIndex]
  returnArray.append(String(substring))

  return returnArray
}

let pieces = splitting("Dog,Cat,Badger,Snake,Lion", delimiter: ",")
print(pieces)

技术点：

• 数组的使用
• indices的使用

练习5

• For example, if the string is "My dog is called Rover" then the resulting string would be "yM god si dellac revoR".

func reversedByWord(sentence: String) -> String {
  var reversedWords = ""
  var lastWordIndex = sentence.startIndex

  for i in sentence.indices {
    if sentence[i] == " " {
      let substring = sentence[lastWordIndex..<i]
      reversedWords += substring.reversed() + " "
      lastWordIndex = sentence.index(after: i)
    }
  }

  // Add the final word
  let substring = sentence[lastWordIndex..<sentence.endIndex]
  reversedWords += substring.reversed()

  return reversedWords
}

let reversed = reversedByWord(sentence: "The quick brown fox jumps over the lazy dog")
print(reversed)

技术点：

• 使用indices遍历字符串索引。
• reversed实现字符串反转。
• substring实现内存优化（和练习4相比）