Question

我正在尝试完成此

a = [5,6,7] b = [3,6,10]

现在，让我们比较每个人的得分：

a[0] > b[0], so Alice receives  point.
a[1] == b[1], so nobody receives a point.
a[2] < b[2] , so Bob receives  point.

Alice的比较得分为1，而Bob的比较得分为1.因此，我们必须在一行上打印1 1（Alice的比较得分，然后是Bob的比较得分）。

我是这样做的：

var a = [5, 6, 7]
var b = [3, 6, 10]

let pointsAlice = (a[0] > b[0] ? 1: 0) + (a[1] > b[1] ? 1: 0) + (a[2] > b[2] ? 1: 0)

let pointsBob = (a[0] < b[0] ? 1: 0) + (a[1] < b[1] ? 1: 0) + (a[2] < b[2] ? 1: 0)

print("\(pointsAlice) \(pointsBob)")

但它显示为不正确的答案（Hackerrank）。我的方式是正确的还是有其他方法可以解决这个问题？

Answer 1

请试试这个......

let a = readLine()!.components(separatedBy: " ").flatMap { Int($0) }
let b = readLine()!.components(separatedBy: " ").flatMap { Int($0) }
let aResult = zip(a, b).filter(>).count
let bResult = zip(b, a).filter(>).count

//Final Result Status 
print(aResult, bResult)

Answer 2

我想你更专注于解释：p 试试这个：

let arrA = readLine()!.components(separatedBy: " ").map{ Int($0)! }
let arrB = readLine()!.components(separatedBy: " ").map{ Int($0)! }

var aliceScore = 0
var bobScore = 0

for index in 0...2{
    if (arrA[index] > arrB[index]){
        aliceScore += 1
    }
    if (arrA[index] < arrB[index]){
        bobScore += 1
    }
}
print("\(aliceScore)" + " " + "\(bobScore)")

Answer 3

为什么不使用像reduce这样的高阶函数呢？

let a = [5, 6, 7]
let b = [3, 6, 10]

let pointsAlice = zip(a, b).reduce(0, combine: { $0 + ($1.0 > $1.1 ? 1 : 0) } )
let pointsBob = zip(b, a).reduce(0, combine: { $0 + ($1.0 > $1.1 ? 1 : 0) } )

print("\(pointsAlice) \(pointsBob)") // 1 1

Answer 4

不要像这样硬编码

(a[0] > b[0] ? 1: 0) + (a[1] > b[1] ? 1: 0) + (a[2] > b[2] ? 1: 0)

想象一下，如果数组有1000个或更多元素!!!

我认为最简单的解决方案是：

var alicePoints = 0
var bobPoints = 0

for i in 0..<A.count {
    if A[i] > B[i] {
        alicePoints += 1
    } else if A[i] < B[i] {
        bobPoints += 1
    }
}

print("\(alicePoints)  \(bobPoints)")

更优雅的解决方案是上面的@koropok（我为Swift 3.x提升了）

let bobPoints = zip(B, A).reduce(0) { $0 + ($1.0 > $1.1 ? 1 : 0) }
let alicePoints = zip(A, B).reduce(0) { $0 + ($1.0 > $1.1 ? 1 : 0) }

print("\(alicePoints)  \(bobPoints)")

Answer 5

这是更新三胞胎的答案

如果我们有

var a = [5, 6, 7]
var b = [3, 6, 10]

然后

// compareTriplets function
func compareTriplets(a: [Int], b: [Int]) -> [Int] {

let bobPoints = zip(b, a).reduce(0) { $0 + ($1.0 > $1.1 ? 1 : 0) }
let alicePoints = zip(a, b).reduce(0) { $0 + ($1.0 > $1.1 ? 1 : 0) }

print("Points earned by Alice and Bob : \(alicePoints)  \(bobPoints)")

return [alicePoints, bobPoints]

}

Answer 6

这是我的解决方案。试图了解reduce函数的工作原理。

func compareTriplets(a: [Int], b: [Int]) -> [Int] {
   var total = [0,0]

   for x in 0..<a.count {
      if a[x] > b[x] {
         total[0] += 1
      } else if a[x] < b[x] {
        total[1] += 1
      }
   }
   return  aTotal   
}

Answer 7

尝试一下： return [zip(a, b).reduce(0) {$0 + ($1.0 > $1.1 ? 1 : 0)}, zip(b, a).reduce(0) {$0 + ($1.0 > $1.1 ? 1 : 0)}]

Answer 8

在python中，不考虑提出的约束（ab >= 1 & ab<=100）：

def compareTriplets(a, b):
    i = 0
    score = [0,0]

    while i < 3:
        if a[i] > b[i]:
            score[0] = score[0] + 1
        elif a[i] < b[i]:
            score[1] = score[1] + 1
        i = i + 1

     return score

Answer 9

import java.util.*.;

public class Solution {
    public static void main(String[] args)   {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int klice=0;
        int bob=0;
        int a[]=new int[n];
        int b[]=new int[n];
        for(int i=0;i<n;i++){
            a[i]=sc.nextInt();
        }
        for(int i=0;i<n;i++){
            b[i]=sc.nextInt();
        }
        for(int i=0;i<n;i++){
            if(a[i]>b[i]){
                klice++;
            } else if(b[i]>a[i]){
                bob++;
            }
        }
        System.out.println(klice +" "+ bob);
    }
}

比较数组[三胞胎]

9 个答案: