从 Swift 中的数组数组中删除较小的值集

Question

给定一个由包含整数的数组组成的数组。

[[2], [3], [2, 2], [5], [7], [2, 2, 2], [3, 3]]

在 Swift 中，删除包含具有特定值的元素数量较少的数组并仅保留包含该值的较大数组的首选方法是什么。

上面输入的结果是

[[5], [7], [2, 2, 2], [3, 3]]

Answer 1

只是使用 forEach 的替代方法：

let arrays = [[2],  [2, 2], [5], [7], [2, 2, 2], [3, 3], [3]]
var largest: [Int: [Int]] = [:]

arrays.forEach({
    guard let first = $0.first else { return }
    largest[first] = [Int](count: max($0.count,largest[first]?.count ?? 0), repeatedValue: first)
})
Array(largest.values)  // [[5], [7], [2, 2, 2], [3, 3]]

Answer 2

使用[Int: [Int]] 字典来跟踪键指定值的最大数组。

let arrays = [[2], [3], [2, 2], [5], [7], [2, 2, 2], [3, 3]]
var largest = [Int: [Int]]()

for arr in arrays {
    // Get the first value from the array
    if let first = arr.first {

        // current is the count for that key already in dictionary largest
        // If the key isn't found, the nil coalescing operator ?? will
        // return the default count of 0.
        let current = largest[first]?.count ?? 0

        // If our new array has a larger count, put it in the dictionary
        if arr.count > current {
            largest[first] = arr
        }
    }
}

// Convert the dictionary's values to an array for the final answer.   
let result = Array(largest.values)

print(result)  // [[5], [7], [2, 2, 2], [3, 3]]

---

同样的逻辑可以与reduce一起使用，在一行中提供结果：

let result = arrays.reduce([Int: [Int]]()) { var d = $0; guard let f = $1.first else { return d }; d[f] = d[f]?.count > $1.count ? d[f] : $1; return d }.map { $1 }

---

替代版本

此版本使用[Int: Int] 字典来保存为每个键找到的最大数组的计数，然后使用数组构造函数在最后重建数组。

let arrays = [[2], [3], [2, 2], [5], [7], [2, 2, 2], [3, 3]]
var counts = [Int: Int]()

for arr in arrays {
    if let first = arr.first {
        counts[first] = max(counts[first] ?? 0, arr.count)
    }
}

let result = counts.map { [Int](count: $1, repeatedValue: $0) }

print(result)  // [[5], [7], [2, 2, 2], [3, 3]]

---

同样的逻辑可以与reduce一起使用，在一行中提供结果：

let result = arrays.reduce([Int: Int]()) { var d = $0; guard let f = $1.first else { return d }; d[f] = max(d[f] ?? 0, $1.count); return d }.map { [Int](count: $1, repeatedValue: $0) }

Answer 3

当我看到 vacawama 的回复非常相似时，我正要写下我的答案。决定回到它只是因为它是一个有趣的问题。因此，我的替代方案几乎可以肯定比 vacawama 的解决方案慢得多，并且不保留顺序，但我认为作为在 Swift 中解决此类问题的替代方案示例，我认为这很有趣。

var items = [[2], [3], [2, 2], [5], [7], [2, 2, 2], [3, 3]]

let reduced = items.sort({
        let lhs = $0.first, rhs = $1.first
        return lhs == rhs ? $0.count > $1.count : lhs < rhs
    }).reduce( [[Int]]()) { (res, items) in
        return res.last?.last != items.last ? res + [items] : res
    }

print(reduced)  // [[2, 2, 2], [3, 3], [5], [7]]

或者，如果您宁愿将所有内容都塞在一行中：

var items = [[2], [3], [2, 2], [5], [7], [2, 2, 2], [3, 3]]

let reduced = items.sort({ let lhs = $0.first, rhs = $1.first; return lhs == rhs ? $0.count > $1.count : lhs < rhs }).reduce([[Int]]()) { $0.last?.last != $1.last ? $0 + [$1] : $0 }

print(reduced)  // [[2, 2, 2], [3, 3], [5], [7]]