使用通用方法根据纸牌游戏规则对数组进行排序

Question

cards = ['杰克', 8, 2, 6, '国王', 5, 3, '皇后', "杰克", "皇后", "国王"] 使用通用方法根据纸牌游戏规则对数组进行排序。

Answer 1

一种方法是使用所有卡片按正确顺序排列的数组作为参考，并根据参考数组中的索引对每张卡片进行排名。

let cards = ['Jack', 8, 2, 6, 'King', 5, 3, 'Queen',"Jack","Queen","King"];

// change this to match the rules of card game
let theRightOrder = ["Ace", 2, 3, 4, 5, 6, 7, 8, 9, 10, "Jack", "Queen", "King"];

cards.sort((a, b) => theRightOrder.indexOf(a) - theRightOrder.indexOf(b));

console.log(cards);

Answer 2

此问题所需的输出必须按升序显示。其他大于 10 的物品（Jack、Queen、King）是有限的，因此我们可以为每个物品附加一个重量。

let cards = ['Jack', 8, 2, 6, 'King', 5, 3,"Queen"];
// change this to match the rules of card game

const otherItems = {
'Jack': 11,
'Queen': 12
'King': 13
}

cards.sort((a, b) => {
if(isNaN(a)) a = otherItems[a];
if(isNaN(b)) b = otherItems[b];
return a-b;
});

console.log(cards);

权重大于 10 的项目可以在

looked up

 上

otherItems

，而不是在数组中查找索引

Answer 3

let cards = ['Jack', 8, 2, 6, 'King', 5, 3, 'Queen', "Jack", 
             "Queen","King"]
const otherItems = {
'Ace':1,
'Jack': 11,
'Queen': 12,
'King': 13
}


const solutionTwo= (someArray)=>{

someArray.sort((a,b)=>{
    a= isNaN(a)? otherItems[a]:a;
    b= isNaN(b)? otherItems[b]:b;
    return a-b;
});

console.log("Solution Two array", someArray);
}

solutionTwo(cards);

Answer 4

这是不相关的语言，但如果有人想使用 python 解决这个问题，这是我在采访中提出的代码，要求使用 3 个不同的测试用例提供相同的解决方案。

def test_case(cards):
    int_list = sorted([x for x in cards if type(x) == int])
    str_list = sorted([x for x in cards if type(x) == str])
    for x in str_list:
        if x == "King":
            str_list.remove(x)
            str_list.append(x)
    sorted_cards = int_list + str_list
    print(f"TEST CASE = {sorted_cards}")

    test_case(['Jack', 8, 2, 6, 'King', 5, 3, 'Queen'])
    test_case(['Jack', 8, 2, 6, 'King', 5, 3, 'Queen', 'Jack', 'King', 'Queen', 'Queen', 'King', 'Jack'])
    test_case(['Jack', 8, 2, 6, 5, 3])

Answer 5

func run() -> Bool {

var cards: [Any] = ["Jack", 8, 2, 6, "King", 5, 3, "Queen","Queen","king"]

let otherItems: [String: Int] = [
    "Jack": 11,
    "Queen": 12,
    "King": 13
]
cards.sort { a, b in
    var aValue = a
    var bValue = b
    
    if let aString = a as? String, let value = otherItems[aString] {
        aValue = value
    }
    
    if let bString = b as? String, let value = otherItems[bString] {
        bValue = value
    }
    
    if let aValue = aValue as? Int, let bValue = bValue as? Int {
        return aValue < bValue
    }
    
    return false
}
return false

}

Answer 6

此解决方案避免了改变原始数组，而是按排序顺序提供了卡片的浅表副本。它允许使用参数来调用来指示是否：

a.

Ace

为低（默认）或高

b.需要删除重复项（默认）或保留

c.排序后的数组需要升序（默认）或降序

const handleDupes = (fl, ar) => (fl ? [...new Set(ar)] : ar);

const sortCards = (arr, aceHigh = false, removeDupes = false, descending = false) => {
  const sorted = [...Array(9)].map(
    (x, i) => (i + 2)
  ).concat(
    "Jack, Queen, King".split(', ')
  );
  if (aceHigh) sorted.push("Ace");
  else sorted.splice(0, 1, "Ace");
  return handleDupes(
    removeDupes, 
    [...arr]
    .sort(
      (a, b) => (
        sorted.indexOf(a) > sorted.indexOf(b)
        ? descending ? -1 : 1
        : descending ? 1 : -1
      )
    )
  );
};

const unsorted = ['Jack', 8, 2, 6, 'King', 5, 3, 'Queen', "Jack", "Queen", "King"];

console.log(sortCards(unsorted, true, true));