Question

给出一串偶数和奇数，找到哪个是唯一的偶数还是唯一的奇数。

示例： detectOutlierValue（“ 2 4 7 8 10”）; // => 2-第三个数字为奇数，其余数字为偶数

即使我已经将所有内容都转换为数字，为什么还要再次解析int（偶数）？

function detectOutlierValue(str) {
  //array of strings into array of numbers
  newArr = str.split(" ").map(x => parseInt(x))

  evens = newArr.filter(num => num % 2 === 0)
  odds = newArr.filter(num => num % 2 !== 0)

  //if the array of evens has just 1 item while the majority is odds, we want to return the position of that item from the original string.

  if (evens.length === 1) {
    return newArr.indexOf(parseInt(evens)) 
  } else {
    return newArr.indexOf(parseInt(odds)) 
  }
}

Answer 1

这是因为evens和odds是将它们放入indexOf时的数组。尝试用每个数组的第一个值替换最后两行：

function detectOutlierValue(str) {
  //array of strings into array of numbers
  newArr = str.split(" ").map(x => parseInt(x))

  evens = newArr.filter(num => num % 2 === 0)
  odds = newArr.filter(num => num % 2 !== 0)

  //if the array of evens has just 1 item while the majority is odds, we want to return the position of that item from the original string.

  if (evens.length === 1) {
    return newArr.indexOf(evens[0]) 
  } else {
    return newArr.indexOf(odds[0]) 
  }
}

Answer 2

原因是evens和odds不是numbers。它们是arrays。在这种情况下，odds = [7]。因此，您parseInt([7])得到7。请在控制台中查看赔率。您返回odds[0]和evens[0]

function detectOutlierValue(str) {
  //array of strings into array of numbers
  newArr = str.split(" ").map(x => parseInt(x))

  evens = newArr.filter(num => num % 2 === 0)
  odds = newArr.filter(num => num % 2 !== 0)
  //if the array of evens has just 1 item while the majority is odds, we want to return the position of that item from the original string.

  if (evens.length === 1) {
    return newArr.indexOf(evens[0]) 
  } else {
    console.log("odds =",odds);
    return newArr.indexOf(odds[0]) 
  }
}
console.log(detectOutlierValue("2 4 7 8 10"))

检测字符串数组中的离群值-VanillaJS和parseInt

2 个答案: