四舍五入到下一个整数 javascript

Question

我想在 JavaScript 中实现这样的目标：

input = 2455.55
f(input) = 2456
f(input) = 2460
f(input) = 2500
f(input) = 3000
f(input) = 2455.55

我目前正在使用 Math.round() 方法，但只能达到 2,546。想知道是否有最好的方法来实现其余部分。

Answer 1

您可以将您的数字除以十，直到得到一个非整数，将其四舍五入，然后在相同的时间内再次乘以十。像这样：

    function roundUp(n) {
    
        var n2 = n;
        var i=0;
        while (Number.isInteger(n2)) {
           n2 /= 10;
            i++;
        }
        return Math.round(n2) * Math.pow(10, i);
    
    }

    console.log(roundUp(2455.55)); // 2456
    console.log(roundUp(2456)); // 2460
    console.log(roundUp(2460)); // 2500
    console.log(roundUp(2500)); // 3000

Answer 2

谢谢，不错！受你的启发，我这样解决了：

function roundNumber(num, n) {
  const divider = Math.pow(10, n);
  return Math.round(num / divider) * divider;
};

Answer 3

根据您想要的输出，您似乎需要跟踪函数调用的次数。这似乎不是您的函数的参数。

鉴于您只有一个参数的限制，实现看起来可能像

var lastNum = 0
var digitsToRound = 0

function roundUp(input) {
  // Verify whether the function is called with the same argument as last call.
  // Note that we cannot compare floating point numbers.
  // See https://dev.to/alldanielscott/how-to-compare-numbers-correctly-in-javascript-1l4i
  if (Math.abs(input - lastNum) < Number.EPSILON) {
    // If the number of digitsToRound exceeds the number of digits in the input we want
    // to reset the number of digitsToRound. Otherwise we increase the digitsToRound.
    if (digitsToRound > (Math.log10(input) - 1)) {
      digitsToRound = 0;
    } else {
      digitsToRound = digitsToRound + 1;
    }
  } else {
    // The function was called with a new input, we reset the digitsToRound
    digitsToRound = 0;
    lastNum = input;
  }

  // Compute the factor by which we need to divide and multiply to round the input
  // as desired.
  var factor = Math.max(1, Math.pow(10, digitsToRound));
  return Math.ceil(input / factor) * factor;
}


console.log(roundUp(2455.55)); // 2456
console.log(roundUp(2455.55)); // 2460
console.log(roundUp(2455.55)); // 2500
console.log(roundUp(2455.55)); // 3000

四舍五入到下一个整数 javascript

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