数组中重复元素的处理

1、取出数组中重复的元素（不重复的不提取，合并重复的元素）

let datas = [1, 2, 3, 3, 2, 1, 1, 1, 2, 3, 4]
let duplicate = Set(datas.filter({ (number) in datas.filter({ $0 == number }).count > 1 }))

//  [2, 3, 1]

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2、合并重复的元素 （不重复的也提取）

let datas = [1, 2, 3, 3, 2, 1, 1, 1, 2, 3, 4]
Set(datas)

//  [2, 4, 3, 1]

---

3、剔除数组中重复的元素

let datas = [1, 2, 3, 3, 2, 1, 1, 1, 2, 3, 4]
let different = Set(datas).filter({ (number) in datas.filter({ $0 == number }).count == 1 })

//  [4]

---

4、将重复的元素提取出来，生成新的数组

let datas = [1, 2, 3, 3, 2, 1, 1, 1, 2, 3, 4]
let newArray = Set(datas).map({ (number) in datas.filter({ $0 == number }) })

//  [[2, 2, 2], [4], [3, 3, 3], [1, 1, 1, 1]]

如果想要保留顺序，那么这么做：

let datas = [3, 1, 2, 3, 3, 2, 1, 1, 1, 2, 3, 4]

var groupped: [[Int]] = []
datas.enumerated().forEach { (index, element) in
    if let index = groupped.index(where: { $0.last == element }) {
        groupped[index].append(element)
    } else {
        groupped.append([element])
    }
}
//  [[3, 3, 3, 3], [1, 1, 1, 1], [2, 2, 2], [4]]