如何遍历字符串类型的元素

Question

我正在编写一个应该返回重复数的函数。我不明白如何遍历字符串类型的元素。

func countDuplicates(_ s:String) -> Int {
    var duplicates = 0

    for i in s {
        duplicates += 1
    }

    return duplicates
}

Answer 1

你可以试试：

func countDuplicates(_ s:String) -> Int {
    return s.count - Set(s).count
}

地点：

s.count // total number of characters
Set(s).count // number of unique characters

Answer 2

您无需通过字符串来查找重复项的数量。您可以只减去计数和称为Set 的不重复版本。

func countDuplicates(_ s: String) -> Int { s.count - Set(s).count) }

---

注意

您应该考虑那里可能会有多个重复字符。因此，您需要找到所有重复项，例如通过对相似的分组然后对它们进行计数：

let duplications = Dictionary(grouping: text, by: {$0}).map { [$0.key: $0.value.count] }
print(duplications)

---

如果你愿意，你可以得到重复的总和：

let totalDuplicationCount = dups.filter {$0.count > 1}.reduce(into: 0) { $0 += $1.count - 1 }
print(totalDuplicationCount)

Answer 3

如果您想查找重复项的总和，下面的代码可能会有所帮助：

let text = "HelloooE"
func numberOfDuplicates(_ s: String) -> Int { s.count - Set(s).count }

print(numberOfDuplicates(text)) //Answer: 3 >> which says one "l" and two "o"s are duplicated.

但如果需要计算重复字符的数量，答案应该是这样的：

let text = "HelloooE"
func countOfDuplicateChars(_ s: String) -> Int { s.reduce(into: [:]) {result, word in result[word, default: 0] += 1 }
                                                  .filter { $0.value > 1 }
                                                  .count }

print(countOfDuplicateChars(text)) //Answer: 2 >> which says "l" and "o" are duplicated characters.

下面的函数也是返回重复字符的计数，并且不区分大小写：

let text = "HelloooE"
func countOfDuplicateChars_CS(_ s: String) -> Int { s.lowercased()
                                                     .reduce(into: [:]) {result, word in result[word, default: 0] += 1 }
                                                     .filter { $0.value > 1 }
                                                     .count }

print(countOfDuplicateChars(text)) //Answer: 3 >> which says "l", "o" and "e" are duplicated characters.