用Javascript将ForLoop转换为ForEach

如何解决用Javascript将ForLoop转换为ForEach

我正在练习Hackerrank JavaScript问题。我找到了一个名为Compare the triplets的测试。这是问题所在：

a = [1,2,3]
b = [3,1]
 For elements *0*,Bob is awarded a point because a[0] .
 For the equal elements a[1] and b[1],no points are earned.
 Finally,for elements 2,a[2] > b[2] so Alice receives a point.
 The return array is [1,1] with Alice's score first and Bob's second.
 

我找到了这样的解决方案：

let a = [17,28,30];
let b = [99,16,8];
function compareTriplets(a,b) {
  let scoreboard = [0,0];
  for (let i = 0; i < a.length; i++) {
    if (a[i] > b[i]) scoreboard[0]++
    else if (a[i] < b[i]) scoreboard[1]++
  }
  return scoreboard
}
compareTriplets(a,b)

我想将ForLoop转换为ForEach方法。但是我找不到做到这一点的方法。

解决方法

let a = [17,28,30];
let b = [99,16,8];
function compareTriplets(a,b) {
  let scoreboard = [0,0];
  a.forEach((element,i) => {
    if (element > b[i]) scoreboard[0]++
    else if (element < b[i]) scoreboard[1]++
  });
  return scoreboard
}
compareTriplets(a,b)

,

这不是您要的，但让我向您展示一些东西：

public static void main(String[] args) {
    String input = "stuff\n blah\n--payload {'meh': 'kar\n'}";
    // Wanted output: Output: "stuff blah --payload {'meh': 'kar\n'}"

    String regexPayload = "--payload\\s[^\\}]+\\}";
    Matcher matcherExtractPayload = Pattern.compile(regexPayload,Pattern.DOTALL).matcher(input);
    Matcher matcherReplaceWithTag = Pattern.compile(regexPayload).matcher(input);

    String tag = "#PAYLOAD#";
    String taggedPayload = "EMPTY";
    String payLoad = "NO_PAYLOAD_FOUND";
    if(matcherExtractPayload.find()) {
        payLoad = matcherExtractPayload.group();
        taggedPayload = matcherReplaceWithTag.replaceFirst(tag);
    }

    String removedNewline = Pattern.compile("\n").matcher(taggedPayload).replaceAll("");
    String restoredPayload = removedNewline.replaceFirst(tag," " + payLoad);

    System.out.println(restoredPayload); // Output: "stuff blah --payload {'meh': 'kar\n'}"
}

或更小的噪音：

function compareTriplets(a,b) {
  return [
    (a[0] > b[0]) + (a[1] > b[1]) + (a[2] > b[2]),(a[0] < b[0]) + (a[1] < b[1]) + (a[2] < b[2])
  ]
}

更简单，更快，更短。

我的意思是，它的字面意思是“比较三胞胎”。没有任何动态长度或任何东西。而且循环很短。您可以轻松unroll the loop。

function compareTriplets([a,b,c],[d,e,f]) {
  return [
    (a > d) + (b > e) + (c > f),(a < d) + (b < e) + (c < f)
  ]
}