问从Swift中的数组中移除较小的值集

EN

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]]

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 }

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]]

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) }

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]]

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]]

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]]