如何比较数组中的3个值与“not”或“all”相等?

( 备注: 以下解决方案针对 集合游戏,三个值都是0、1或2。不会有帮助的 检查是否有三个 任意的 值都相等或都不同。)

一点数学知识有帮助。如果有三个整数x,y,z在范围内 0到2,那么x+y+z是3的倍数if(并且仅当)

• x,y,z都相等,或者
• x,y,z都是不同的。

这个可以查一下

if (x + y + z) % 3 == 0 {
    // x, y, z are all equal or all different.
}

因此,如果 Card 类型定义为

struct Card {
    let number: Int // 0, 1, 2
    let symbol: Int // 0, 1, 2
    let shade:  Int // 0, 1, 2
    let color:  Int // 0, 1, 2
}

然后,对有效集的检查可以非常有效地完成:

func isValidMatch(c1: Card, c2: Card, c3: Card) -> Bool {
    return (c1.number + c2.number + c3.number) % 3 == 0
        && (c1.symbol + c2.symbol + c3.symbol) % 3 == 0
        && (c1.shade + c2.shade + c3.shade) % 3 == 0
        && (c1.color + c2.color + c3.color) % 3 == 0
}

或者,为各种属性定义原始值在0、1、2范围内的枚举:

struct Card {

    enum Number: Int {
        case one, two, three
    }

    enum Symbol: Int {
        case diamond, squiggle, oval
    }

    enum Shade: Int {
        case solid, striped, open
    }

    enum Color: Int {
        case red, green, purple
    }

    let number: Number
    let symbol: Symbol
    let shade: Shade
    let color: Color
}


func isValidMatch(c1: Card, c2: Card, c3: Card) -> Bool {
    return (c1.number.rawValue + c2.number.rawValue + c3.number.rawValue) % 3 == 0
        && (c1.symbol.rawValue + c2.symbol.rawValue + c3.symbol.rawValue) % 3 == 0
        && (c1.shade.rawValue + c2.shade.rawValue + c3.shade.rawValue) % 3 == 0
        && (c1.color.rawValue + c2.color.rawValue + c3.color.rawValue) % 3 == 0
}

你的 Card 类应符合 Equatable 协议:

class Card : Equatable {
    let color: Int
    let shape: Int
    let shade: Int
    let number: Int

    //This is a failable initializer (https://developer.apple.com/swift/blog/?id=17) since we haven't used enums like in another answer, which are safer. 
    init?(color: Int, shape: Int, shade: Int, number: Int) {
        //make sure that all properties are between 0 and 2
        guard 0...2 ~= color, 0...2 ~= shape, 0...2 ~= shade, 0...2 ~= number else {
            fatalError("All properties should be integers between 0 and 2")
        }
        self.color = color
        self.shape = shape
        self.shade = shade
        self.number = number
    }

    static func == (lhs: Card, rhs: Card) -> Bool {
        return lhs.color == rhs.color && lhs.shape == rhs.shape && lhs.shade == rhs.shade && lhs.number == rhs.number
    }
}

所以当所有四个属性都相等时,就有了相等。如所指出的 @ MartinR :在Swift 4.1及以上版本中,编译器将合成 == 操作员自动地,并且足以声明符合等价协议。

为了符合等价协议,您所要做的就是实现这个功能:

static func == (lhs: Card, rhs: Card) -> Bool

现在,如果要检查三张牌是否相等,请使用 = 操作员:

let c1 = Card(color: 2, shape: 1, shade: 0, number: 1)
let c2 = Card(color: 2, shape: 1, shade: 0, number: 1)
let c3 = Card(color: 2, shape: 1, shade: 0, number: 1)
let c4 = Card(color: 2, shape: 1, shade: 0, number: 2)

(c1 == c2) && (c2 == c3)

最后,如果您想要一个函数一次比较三张卡,可以将此代码添加到 卡 班级:

class func compareThreeCards(firstCard: Card, secondCard: Card, thirdCard: Card) -> Bool {
    return (firstCard == secondCard) && (secondCard == thirdCard)
}

你可以这样使用它:

Card.compareThreeCards(firstCard: c1!, secondCard: c2!, thirdCard: c3!)

自Swift4.1以来,结构和枚举可以免费拥有相等的和可哈希的(不适用于类)。

例如,如果您有一个卡片结构,只需使这个类型是可哈希的,用您的卡片创建一个集合,然后比较它们。

struct Card: Hashable {

    var color: Int
    var shape: Int
    var shade: Int
    var number: Int
}

let card1 = Card(color: 2, shape: 2, shade: 2, number: 0)
let card2 = Card(color: 2, shape: 2, shade: 2, number: 1)
let card3 = Card(color: 3, shape: 1, shade: 0, number: 2)


print(card1 == card2 && card2 == card3) // Prints false, all different.

let card4 = Card(color: 1, shape: 2, shade: 0, number: 0)
let card5 = Card(color: 1, shape: 2, shade: 0, number: 0)
let card6 = Card(color: 1, shape: 2, shade: 0, number: 0)

print(card4 == card5 && card5 == card6) // Prints true, all the same.

注(感谢利奥·达布斯):布景是一个无序的收藏。