如何摆脱算法中的复杂性? [英] How to get rid of the complexity in the algorithm?

问题描述

练习 编写一个乘数 (a, b) 函数，将数字 a 与数字 b 相乘，而不使用*"运算符或 Math.imul 方法.

Exercise: Write a multiple (a, b) function that multiplies the number a by the number b without using the "*" operator or the Math.imul method.

 multiple(1, 1) // 1
 multiple(1, 2) // 2
 multiple(0, 0) // 0

代码:

export default function multiple(a, b) {
  if (a === Infinity && b === 0 || a === -Infinity && b === 0 || a === 0 && b === Infinity || a === 0 && b === -Infinity) {
    return NaN;
  }

  if (b === Infinity) {
    if (a < 0) {
      return -Infinity;
    }
    return Infinity;
  } if (b === -Infinity) {
    if (a < 0) {
      return Infinity;
    }
    return -Infinity;
  }

  if (b === 1 || b === -1) {
    if (a < 0 && b < 0 || a > 0 && b < 0) {
      return -a;
    }
    return a;
  }

  if (b < 0) {
    return -a + multiple(a, b + 1);
  }
  return a + multiple(a, b - 1);
}

好的，我写了这段代码，它通过了测试.但是 eslint 抱怨函数复杂度过高:Function 'multiple' 的复杂度为 15.最大允许为 10.eslint(复杂度)

Ok, I write this code and it passes the tests. But eslint complains about over complexity function: Function 'multiple' has a complexity of 15. Maximum allowed is 10. eslint (complexity)

如何降低函数的复杂度?

更新

const multiple = (a, b) => a / (1 / b);

是的，它确实有效，但是如果查看我的代码，我可以缩短重复操作的时间，我看起来很盲目，但我想了解它.

Yes its really worked, but if look on my code, where I can short a repeated operations, I looks blind, but I want to understand it.

更新 2

解决方案必须经过所有测试:

Solution must goes all tests:

const random = () => Math.floor(Math.random() * 100) * (Math.random() < 0.5 ? -1 : 1)

const cases = [
    [0, 1],
    [1, 0],
    [1, 1],
    [1, 2],
    [0, 0],
    [5, 5],
    [5, -5],
    [290, -41],
    [-5, 5],
    [-5, -5],
    [random(), random()],
    [random(), random()],
    [random(), random()],
    [random(), random()],
    [random(), random()],
    [10, -Infinity],
    [10, Infinity],
    [-10, Infinity],
    [-10, -Infinity],
    [Infinity, 10],
    [-Infinity, -10],
    [Infinity, -10],
    [-Infinity, 10],
    [0, Infinity],
    [0, -Infinity],
    [Infinity, 0],
    [-Infinity, 0]
]

cases.forEach(([a, b]) => {
    console.log(`\na:${a} b:${b}`)
    console.log('my:', multiple(a, b), 'Fact:', a * b)
})

推荐答案

您可以减少测试标志的测试次数.您可以改为将带有负数的调用转换为带有正数的调用并否定结果.

You can reduce the number of tests where you test the sign. You could instead translate the call with a negative number to a call with a positive number and negate the result.

这是一个更紧凑的版本.

Here is a more compact version.

注意:我认为练习的精神是也不使用 /，否则使用 a*b === a/(1/b)

NB: I assume the spirit of the exercise was to not use / either, as otherwise it is trivial to use the fact that a*b === a/(1/b)

function multiple(a, b) {
  return a < b ? multiple(b, a)
       : b < 0 ? -multiple(a, -b)
       : a === Infinity ? (b ? a : NaN)
       : b === Infinity || !b ? b
       : a + multiple(a, b - 1);
}

// Tests:

const random = () => Math.floor(Math.random() * 200) - 100;

const cases = [
    [0, 1],
    [1, 0],
    [1, 1],
    [1, 2],
    [0, 0],
    [5, 5],
    [5, -5],
    [290, -41],
    [-5, 5],
    [-5, -5],
    [random(), random()],
    [random(), random()],
    [random(), random()],
    [random(), random()],
    [random(), random()],
    [10, -Infinity],
    [10, Infinity],
    [-10, Infinity],
    [-10, -Infinity],
    [Infinity, 10],
    [-Infinity, -10],
    [Infinity, -10],
    [-Infinity, 10],
    [0, Infinity],
    [0, -Infinity],
    [Infinity, 0],
    [-Infinity, 0]
]

cases.forEach(([a, b]) => {
    let result = multiple(a, b);
    if (!Object.is(result, a*b)) {
        console.log(`\na:${a} b:${b} my result: ${result}, expected: ${a*b}`);
    }
})
console.log("all done");

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