arrays - 从 Swift 中的数组数组中删除较小的值集

Question

给定一个数组，该数组由包含整数的数组组成。

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

在 Swift 中，移除包含具有特定值的较少元素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

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 一起使用以在一行中提供结果:

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] 字典仅保留为每个键找到的最大数组的计数，然后在末尾使用数组构造函数重建数组。

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 一起使用以在一行中提供结果:

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) }