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在 JavaScript 中考慮運算子優先級評估數學表達式

WBOY
WBOY轉載
2023-08-24 15:53:091052瀏覽

在 JavaScript 中考虑运算符优先级评估数学表达式

問題

我們需要寫一個 JavaScript 函數,它將數學表達式作為字串接收,並將其結果作為數字傳回。

我們需要支援以下數學運算子-

  • 除法/(當浮點除法)

  • p>

  • #減法-

  • 乘法*

運算子總是根據從左到右,* 和/ 必須在和- 之前計算。

範例

以下是程式碼 -

 即時示範

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 !== &#39;&#39;) {
         outputQueue.push(new Number(str));
         str = &#39;&#39;;
      }
      if (ch === &#39;-&#39;) {
         if (ind == 0) {
            ch = &#39;negate&#39;;
         } else {
            let nextCh = exp[ind+1];
            let prevCh = exp[ind-1];
            if ((digits.includes(nextCh) || nextCh === &#39;(&#39; || nextCh === &#39;-&#39;) &&
               (operators.includes(prevCh) || exp[ind-1] === &#39;(&#39;)) {
                  ch = &#39;negate&#39;;
            }
         }
      }
      if (operations.length > 0) {
         let topOper = operations[operations.length - 1];
         while (operations.length > 0 && legend[topOper] &&
         ((legend[ch].assoc === &#39;left&#39; && legend[ch].pred <= legend[topOper].pred) ||
         (legend[ch].assoc === &#39;right&#39; && 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 === &#39;(&#39;) {
      operations.push(ch);
   } else if (ch === &#39;)&#39;) {
      if (str !== &#39;&#39;) {
         outputQueue.push(new Number(str));
         str = &#39;&#39;;
      }
      while (operations.length > 0 && operations[operations.length - 1] !== &#39;(&#39;) {
         outputQueue.push(operations.pop());
      }
      if (operations.length > 0) { operations.pop(); }
   }
   ind++;
}
if (str !== &#39;&#39;) { 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 === &#39;negate&#39;) {
            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));

輸出

2

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