Javascript按两个数字序列对数组中的数字求和

Javascript sum numbers in an array by two digit sequence each

本文关键字：数字 数组 求和 字序 Javascript 两个      更新时间：2023-09-26

我被这个练习困住了，最后我告诉你我的输出是什么，但在练习的描述之前，提前感谢!

描述:

它接收一个包含数字和字母的数组，并将其数字美化后返回。字母保持不变美化过程是将一个数字的所有数字相加，使其变成一个个位数:

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']

my CODE: 

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'])

我收到的输出是=> [2, 3, 5, 9, 4, "A", "b"]

这是做总和之前的"前一步"还是我做错了什么?

嗨，你可以像我在评论中提到的那样尝试。

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。

我修改了你的代码。请找到下面的代码，这是根据您的需要给出的结果。

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'])

可以用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))
})

试试:

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'])

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;
 });
}

在您的示例中，您应该将部分更改为:

: 

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

: 

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

没有任何字符串操作的纯函数，这将是我的解决方案;

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));

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