javascript - 仅用减法进行递归除法——无法处理小数

Question

我正在练习我的数学/算法技能，并尝试仅使用减法来除两个数字。我非常接近，但我似乎无法处理小数，我不知道为什么？在堆栈的最低级别，当我调用divide(9,2)时，我注意到我返回“0”，而实际上我想返回1/2 - 但不使用除法运算符...如果 x 小于 y 检查，我应该在子例程中添加该逻辑吗？我陷入了如何在小数点后递归添加最多三位数字的困境。

    var divide = function(x, y) {
        //the number of times you need to subtract y from x.

      if (y === 0) {
        return 0
      } 
      // if 
      if (x - y === 0) {
        return 1;
      } 
      if (x < y) {
    // if this is the case, get the value of y - x.  ->1
    var diff = y - x;
    console.log(diff);
    // add a zero to the end --> so in our case, 10
    diff = String(diff) + '0';
    console.log(diff);
    diff = Number(diff);
    console.log(diff);
    // is that now divisible by y? is so how many times? in our case, 5 times.
    var decimal = Number(divide(diff, y));
    var number = "." + decimal;
    //so add .5 to it.
    return number;
  } else {
        return (1 + divide(x - y, y)); 
      }

    };

    var modulo = function(x, y) {

      var val = x;
      while (val >= y) {
        val -= y;

      }
      return val;

    };

Answer 1

你会认为这是作弊吗？没有除法或乘法，加法已转换为减法，它返回一个 float ，并且使用递归。但是，需要进行大量的字符串转换，以连接数字并添加符号和小数点。

function divide(x, y, prec) {
    if (y == 0) return NaN;
    var quot = 0, sign = 1;
    if (x < 0) { 
        sign = 0 - sign;
        x = 0 - x;
    }
    if (y < 0) {
        sign = 0 - sign;
        y = 0 - y;
    }
    while (x >= y) {
        x -= y;
        quot = 0 - (0 - 1 - quot);
    }
    quot = (sign < 0 ? "-" : "").concat(quot, prec == undefined && x > 0 ? "." : "");
    if (x > 0) {
        prec = prec || 13;
        if (--prec) {
            for (var i = 9, temp = x; i; i--) x = 0 - (0 - x - temp);
            quot = quot.concat(divide(x, y, prec));
        }
    }
    return parseFloat(quot);
}

alert(divide( 9,  2));
alert(divide( 2,  9));
alert(divide( 9, -2));
alert(divide(-2,  9));
alert(divide(-9, -2));
alert(divide( 0,  9));
alert(divide( 9,  0));