javascript - 在javascript中将数字10.0反转为0.01

Question

我正在尝试实现一个在 JavaScript 中反转给定数字(int/float)的函数。函数实现为

function reverseNumber(n) {
    let reversed = 0;
    const sign = Math.sign(n)
    n *= sign;

    const exponent = n.toString().indexOf('.');
    if (exponent !== -1) n *= Math.pow(10, n.toString().length - 1 - exponent);
    while (n !== 0) {
        reversed *= 10;
        reversed += n % 10;
        n = Math.floor(n / 10);
    }
    if (exponent !== -1) reversed /= Math.pow(10, exponent);
    return reversed * sign;
}

该函数通过了以下测试用例

  expect(reverseNumber(0)).toEqual(0);
  expect(reverseNumber(15)).toEqual(51);
  expect(reverseNumber(90)).toEqual(9);
  expect(reverseNumber(-5)).toEqual(-5);
  expect(reverseNumber(-90)).toEqual(-9);
  expect(reverseNumber(-2359)).toEqual(-9532);
  expect(reverseNumber(12.403)).toEqual(304.21);
  expect(reverseNumber(-12.4)).toEqual(-4.21);

唯一失败的情况是输入为 10.0 时。测试用例是

expect(reverseNumber(10.0)).toEqual(0.01);

我没有找到任何方法来反转 10.0 => 0.01。你们能帮忙吗？谢谢

Answer 1

既然您似乎很乐意在字符串级别工作，最简单的事情可能就是转换为字符串，将其转换为数组，反转数组，然后将它们全部放回一起 - 同时允许使用减号:

function reverseNumber(num) {
    // Remember if it's negative
    var isNegative = num < 0;
    // Get the digits and decimal just for the absolute part
    var chars = (isNegative ? -num : num).toString().split("");
    // Reverse them
    chars.reverse();
    // Put it back together and convert back to a number
    var num = Number(chars.join(""));
    // Return the result, converting back to negative if appropriate
    return isNegative ? -num : num;
}

对于教学来说，这很冗长，当然，其中许多操作可以链接在一起。

您的测试示例:

function reverseNumber(num) {
    // Remember if it's negative
    var isNegative = num < 0;
    // Get the digits and decimal just for the absolute part
    var chars = (isNegative ? -num : num).toString().split("");
    // Reverse them
    chars.reverse();
    // Put it back together and convert back to a number
    var num = Number(chars.join(""));
    // Return the result, converting back to negative if appropriate
    return isNegative ? -num : num;
}

function test(num, expect) {
    var result = reverseNumber(num);
    console.log(num, " => ", result, result == expect ? "OK" : "ERROR");
}

test(0, 0);
test(15, 51);
test(90, 9);
test(-5, -5);
test(-90, -9);
test(-2359, -9532);
test(12.403, 304.21);
test(-12.4, -4.21);

.as-console-wrapper {
  max-height: 100% !important;
}

<小时/>

重新编辑:

The only case its failing is when input is 10.0 . test case is

expect(reverseNumber(10.0)).toEqual(0.01);

I am not finding any way to reverse 10.0 => 0.01.

如果您的起点是一个数字，那么您就不能这样做，否则会弄乱其他几个测试用例。作为数字，10 和 10.0 之间没有区别，因此如果 10.0(即 10)=> 0.01，然后 15(即 15.0)=> 0.51)。但你说过它应该是 51。

如果您的起点是字符串，您就可以做到这一点，但是:

function reverseNumber(numStr) {
    // Remember if it's negative
    var isNegative = numStr[0] === "-";
    // Get the digits and decimal just for the absolute part
    var chars = (isNegative ? numStr.substring(1) : numStr).split("");
    // Reverse them
    chars.reverse();
    // Put it back together and convert back to a number
    var num = Number(chars.join(""));
    // Return the result, converting back to negative if appropriate
    return String(isNegative ? -num : num);
}

function test(num, expect) {
    var result = reverseNumber(num);
    console.log(num, " => ", result, result == expect ? "OK" : "ERROR");
}

test("0", "0");
test("15", "51");
test("90", "9");
test("-5", "-5");
test("-90", "-9");
test("-2359", "-9532");
test("12.403", "304.21");
test("-12.4", "-4.21");
test("10.0", "0.01");

.as-console-wrapper {
  max-height: 100% !important;
}