Question

我有这个问题：

编写一个简单的解释器，该解释器可以理解“ +”，“-”和“ *”操作。按顺序应用操作使用以value初始值开头的命令/参数对。如果遇到未知命令，请返回-1。您将完全忽略B.O.D.M.A.S。

Examples of the input and output
interpret(1, ["+"], [1]) → 2
interpret(4, ["-"], [2]) → 2
interpret(1, ["+", "*"], [1, 3]) → 6
interpret(5, ["+", "*", "-"], [4, 1, 3]) → 6

我尝试将参数作为下面的多维数组传递。我正在尝试解决问题，以便当我这样做时

let operator = ["+", "-"];
let integer = 1;
let intArr = [1, 2];
let emptyInt;
let anotherInt;
let newArray = [integer, operator, intArr];

我如何像上面那样进行这项工作？依次添加每个数组

Answer 1

您可以使用Array.prototype.reduce()。ac作为第一个值。然后通过检查运算符来加/减/除/乘。

function interpret(...args){
  let operators = args[1]; //get the operators array
  let values = args[2] //numbers expect the first one.
  return values.reduce((ac,val,i) =>{
    //check the operator at the 'i' on which we are on while iterating through 'value'
    if(operators[i] === '+') return ac + val;
    else if(operators[i] === '-') return ac - val;
    else if(operators[i] === '*') return ac * val;
    else if(operators[i] === '/') return ac / val;
    else return -1;
  },args[0]) //'ac' is initially set to first value.
}

console.log(interpret(1, ["+"], [1])) 
console.log(interpret(4, ["-"], [2])) 
console.log(interpret(1, ["+", "*"], [1, 3])) 
console.log(interpret(5, ["+", "*", "-"], [4, 1, 3]))

Answer 2

您可以使用recursive approach。首先，您可以定义一个对象以将您的运算符映射到可用函数，然后调用递归函数来计算结果：

const oper = {
  '+': (a, b) => a + b,
  '-': (a, b) => a - b,
  '*': (a, b) => a * b,
  '/': (a, b) => a / b
};

const interpret = (n, [fc, ...calcs], [fn, ...nums]) =>
  fc === undefined ? n :
  interpret(oper[fc](n, fn), calcs, nums)

console.log(interpret(1, ["+"], [1])); // 2
console.log(interpret(4, ["-"], [2])); // 2
console.log(interpret(1, ["+", "*"], [1, 3])); // 6
console.log(interpret(5, ["+", "*", "-"], [4, 1, 3])); // 6

如果必须为无效的操作数返回-1，则可以使用以下命令：

const oper = {
  '+': (a, b) => a + b,
  '-': (a, b) => a - b,
  '*': (a, b) => a * b,
  '/': (a, b) => a / b
};

const interpret = (n, [fc, ...calcs], [fn, ...nums]) => {
  if(fc === undefined) return n;
  if(!(fc in oper)) return -1;
  return interpret(oper[fc](n, fn), calcs, nums)
}

console.log(interpret(1, ["+"], [1])); // 2
console.log(interpret(4, ["-"], [2])); // 2
console.log(interpret(1, ["+", "*"], [1, 3])); // 6
console.log(interpret(5, ["+", "*", "-"], [4, 1, 3])); // 6
console.log(interpret(1, ["+", "%"], [1, 2])); // -1

Answer 3

您可以使用具有运算符所需功能的对象，并通过获取值和操作数来重新分配数组。

const
    fns = {
        '+': (a, b) => a + b,
        '-': (a, b) => a - b,
        '*': (a, b) => a * b,
        '/': (a, b) => a / b
    },
    interpret = (start, operators, values) =>
        values.reduce(
            (a, b, i) => operators[i] in fns
                ? fns[operators[i]](a, b)
                : - 1,
            start
        );

console.log(interpret(1, ["+"], [1])); // 2
console.log(interpret(4, ["-"], [2])); // 2
console.log(interpret(1, ["+", "*"], [1, 3])); // 6
console.log(interpret(5, ["+", "*", "-"], [4, 1, 3])); // 6

Answer 4

您可以这样：-

function getOperator(number, operator, integer) {
    let result;
    switch (operator) {
        case '+':
            result = number + integer;
            break;
        case '-':
            result = number - integer;
            break;
        case '*':
            result = number * integer;
            break;
        case '/':
            result = number / integer;
            break;
    }
    return result;
}
function doOperation(array) {
    number = array[0];
    for (let i = 0; i < array[1].length; i++) {
        number = getOperator(number, array[1][i], array[2][i]);
    }
    return number;
}

然后您可以使用它来实现询问：-

doOperation([1, ["+"], [1]])

识别运算符并执行计算

4 个答案: