Question

我坚持这个练习，最后我告诉你我的输出是什么，但在练习描述之前，请提前感谢！

描述：

它接收一个带有数字和字母的数组，并返回它的美化数字。信件保持不变美化过程是通过将所有数字加在一起将数字减少为单个数字：

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：

＆＃13;

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'])

＆＃13;

我收到的输出是=＆gt; [2, 3, 5, 9, 4, "A", "b"]

这是＆＃34;之前的＆＃34;在做总结之前还是我做错了什么？

Answer 1

嗨，你可以像我在评论中提到的那样尝试。

＆＃13;

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']))

＆＃13;

编辑：谢谢@georg的建议。

Answer 2

我已对您的代码进行了更改。请找到下面的代码，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))
})

Answer 4

试试这样：

function beautifyNumbers(array) {

  var newArray = [];
  array.forEach(function(element) {
    // Checks if character is a letter and not a number
    if (typeof element == "number") {
      newArray.push(beautifyElement(element));
    } else {
      // pushes numbers to the array without making
      // any change to them
      newArray.push(element);
    }
  });
  // returns the array
  return newArray;


}

   function beautifyElement(element){
    var sNumber = element.toString();
    var sum = 0;
    for (var i = 0, len = sNumber.length; i < len; i += 1) {
        sum = sum + +sNumber[i];
    }
    if(sum>9){
        return beautifyElement(sum);
    }else{
        return sum;
    }
}

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

Answer 5

在您的示例中，您应该使用以下内容更改部件：

发件人：

for (var i = 0, len = sNumber.length; i < len; i += 1) { newArray.push(+sNumber.charAt(i)); } console.log(newArray);;

以

var mySum = 0; for (var i = 0, len = sNumber.length; i < len; i += 1) { mySum += sNumber.charAt(i); } newArray.push(mySum); console.log(newArray);

Answer 6

function bsum(v) {
    var res = (v+"").split('').reduce(function(a, b) {return (a | 0) + (b | 0);});
    if(res | 0 > 10) return bsum(res) ;
    else return res;
}

function beautifyNumbers(array) {
 return array.map(function(v, i) {
    if (typeof v == 'number') {
        return bsum(v) | 0;
    }
     else return v;
 });
}

Answer 7

纯函数没有任何字符串操作，这将是我的解决方案;

＆＃13;

function funnySum(n){
  var d = Math.floor(Math.log10(n) +1),  // get number of digits
    sum = Array(d).fill()                // prepare the array and place digits
                  .map((_,i) => Math.floor(n % Math.pow(10,d-i) / Math.pow(10,d-1-i)))
                  .reduce((p,c) => p+c); // get sum of array items
  return sum > 9 ? funnySum(sum) : sum;  // if still two digits continue...
}

console.log(funnySum(9956));

＆＃13;

Javascript数组中的数字，每个数字由两位数字组成

7 个答案:

递归