考虑 JavaScript 中的运算符优先级评估数学表达式
问题
我们需要编写一个JavaScript函数，它接受一个数学表达式作为一个字符串，并将其结果作为一个数字返回。
我们需要支持以下数学运算符-
除法/（作为浮点除法）
加法+
减法-
乘法*
运算符总是从左到右计算，*和/必须在+和-之前计算。
示例
以下是代码-
const exp = '6 - 4';
const findResult = (exp = '') => {
const digits = '0123456789.';
const operators = ['+', '-', '*', '/', 'negate'];
const legend = {
'+': { pred: 2, func: (a, b) => { return a + b; }, assoc: "left" },
'-'&: { pred: 2, func: (a, b) => { return a - b; }, assoc: "left" },
'*': { pred: 3, func: (a, b) => { return a * b; }, assoc: "left" },
'/': { pred: 3, func: (a, b) => {
if (b != 0) { return a / b; } else { return 0; }
}
}, assoc: "left",
'negate': { pred: 4, func: (a) => { return -1 * a; }, assoc: "right" }
};
exp = exp.replace(/\s/g, '');
let operations = [];
let outputQueue = [];
let ind = 0;
let str = '';
while (ind < exp.length) {
let ch = exp[ind];
if (operators.includes(ch)) {
if (str !== '') {
outputQueue.push(new Number(str));
str = '';
}
if (ch === '-') {
if (ind == 0) {
ch = 'negate';
} else {
let nextCh = exp[ind+1];
let prevCh = exp[ind-1];
if ((digits.includes(nextCh) || nextCh === '(' || nextCh === '-') &&
(operators.includes(prevCh) || exp[ind-1] === '(')) {
ch = 'negate';
}
}
}
if (operations.length > 0) {
let topOper = operations[operations.length - 1];
while (operations.length > 0 && legend[topOper] &&
((legend[ch].assoc === 'left' && legend[ch].pred <= legend[topOper].pred) ||
(legend[ch].assoc === 'right' && legend[ch].pred < legend[topOper].pred))) {
outputQueue.push(operations.pop());
topOper = operations[operations.length - 1];
}
}
operations.push(ch);
} else if (digits.includes(ch)) {
str += ch
} else if (ch === '(') {
operations.push(ch);
} else if (ch === ')') {
if (str !== '') {
outputQueue.push(new Number(str));
str = '';
}
while (operations.length > 0 && operations[operations.length - 1] !== '(') {
outputQueue.push(operations.pop());
}
if (operations.length > 0) { operations.pop(); }
}
ind++;
}
if (str !== '') { outputQueue.push(new Number(str)); }
outputQueue = outputQueue.concat(operations.reverse())
let res = [];
while (outputQueue.length > 0) {
let ch = outputQueue.shift();
if (operators.includes(ch)) {
let num1, num2, subResult;
if (ch === 'negate') {
res.push(legend[ch].func(res.pop()));
} else {
let [num2, num1] = [res.pop(), res.pop()];
res.push(legend[ch].func(num1, num2));
}
} else {
res.push(ch);
}
}
return res.pop().valueOf();
};
console.log(findResult(exp));输出结果2