Question

我正在Swift中创建一个邻接表，存储一个节点数组。但是，当从现有节点添加边时，我需要检查from密钥是否存在于任何子节点中，并且是否确实要检查to值是否存在于同一子节点中。

func addEdge(from: String, to: String) {
//the full code for addEdge is incomplete here
    if (children.contains{ $0.nodes[from] != nil}) {
    for child in children {
        if (child.nodes[from] != nil) {
            if (!(child.nodes[from]?.contains{$0 == to})!){
                child.nodes[from]?.append(to)
            }
        }
    }
    }
    }

孩子是

var children = [Node]()

并且节点是

class Node: Hashable {
    var nodes = [String:[String]]()
    var hashValue: Int{ return nodes.hashValue }
    static func == (lhs: Node, rhs: Node) -> Bool {
        return lhs.nodes.keys == rhs.nodes.keys
    }
}

现在可以了，但是看起来真的很丑。 Swift中肯定有更好的方法，但这是什么？

Answer 1

假设您不希望更改实现上述代码的方式，但希望提高可读性，则可以利用if let和可选链接来使代码更整洁，更具可读性。

func addEdge(from: String, to: String) {
//the full code for addEdge is incomplete here
    if children.contains{ $0.nodes[from] != nil } {
        for child in children {
            if let fromNode = child.nodes[from], fromNode.contains{$0 == to} {
                fromNode.append(to)
            }
        }
    }
}

Swift Optional Chaining

Answer 2

尝试类似的东西：

if (children.contains{ $0.nodes[from] != nil}) {
    children.filter { $0.nodes[from] != nil }.
       compactMap { $0.nodes[from] }.
       filter { !($0.nodes[from]!.contains{$0 == to}) }.
       forEach { $0.nodes[from]?.append(to) }
}

解开字典值Swift

2 个答案: