数组中的Javascript和数字，每个数字序列由两位数组成

Question

我被这个练习困住了，最后我告诉你我的输出是什么，但是在对这个练习的描述之前，谢谢！

描述：

它接收一个包含数字和字母的数组，并将其返回并美化它的数字。字母保持不变，美化过程是通过将所有数字相加在一起，将一个数字减少为一个单数数字：

123 = 6 because 1+2+3 = 6
9 = 9
9956 = 2 because 9+9+5+6 = 29 -> 2+9 = 11 -> 1+1 = 2
793 = 1 because 7+9+3 = 19 -> 1+9 = 10 -> 1+0 = 1
Example: beautifyNumbers([23,59, 4,'A','b']) returns [5, 5, 4, 'A', 'b']

我的代码：

​

function beautifyNumbers(array) {

  var newArray = [];
  array.forEach(function(element) {
    // Checks if character is a letter and not a number
    if (typeof element == "number") {
      var sNumber = element.toString();

      for (var i = 0, len = sNumber.length; i < len; i += 1) {
        newArray.push(+sNumber.charAt(i));
        // The "+" sign converts a String variable to a Number, if possible: +'21.2' equals Number(21.2).
        // If the conversion fails, it return NaN.
        // El método charAt() de String devuelve el carácter especificado de una cadena:
        // var name="Brave new world"; name.charAt(0) => 'B'
      }
      console.log(newArray);;
    } else {
      // pushes numbers to the array without making
      // any change to them
      newArray.push(element);
    }
  });
  // returns the array
  return newArray;


}

beautifyNumbers([23, 59, 4, 'A', 'b'])

​

A我收到的输出是=> [2, 3, 5, 9, 4, "A", "b"]

这是做之和之前的“前一步”，还是我做错了什么？

Answer 1

嗨，你可以试着像我在评论中提到的那样。

​

function beautifyNumbers(array) {
  var newArray = [];
  array.forEach(function(element) {
    // Checks if character is a letter and not a number
    if (typeof element == "number") {
      if(element %9 == 0 && element != 0)
        newArray.push(9);
      else 
        newArray.push(element%9);
    } else {
      newArray.push(element);
    }
  });
  return newArray;
}
console.log(beautifyNumbers([1231, 0, 18, 27, 12354, 59, 4, 'A', 'b']))

​

编辑:谢谢@georg的建议。

Answer 2

我已经修改了你的代码。请找到下面的代码，这是根据您的需要提供结果。

function beautifyNumbers(array) {
            var newArray = [];
            array.forEach(function (element) {
                // Checks if character is a letter and not a number
                if (typeof element == "number") {
                    var sNumber = element.toString();
                    var beutifySum = 0;
                    for (var i = 0, len = sNumber.length; i < len; i += 1) {
                        beutifySum += +sNumber.charAt(i);
                    }

                    beutifySum = beutifySum % 9 === 0 ? 9 : beutifySum % 9;
                    newArray.push(beutifySum);
                } else {
                    // pushes numbers to the array without making
                    // any change to them
                    newArray.push(element);
                }
            });

            console.log(newArray);
            // returns the array
            return newArray;
        }

beautifyNumbers([23, 59, 4, 'A', 'b'])

Answer 3

您可以使用recursion + array.reduce

递归

​

function sumOfDigits(num) {
  // Check if value is number or alphanumeric
  if (!isNaN(num)) {
    // Convert number and split to get individual values.
    // Loop over values and add them
    var sum = num.toString().split("").reduce(function(p, c) {
      // +p is a shorthand for parseInt(p)
      return +p + +c;
    });

    // check if number is greater than 10. If yes, repeat the process
    if (sum >= 10) sum = sumOfDigits(sum);
    return sum;
  } 
  // if value is not number, return value
  else return num;
}

var data = [123, 4, 567, 'a', "abc", 0];
data.forEach(function(n) {
  console.log(sumOfDigits(n))
})

​