1. var calculate = function(s) {
    2.     s = s.trim();
    3.     const stack = new Array();
    4.     let preSign = '+';
    5.     let num = 0;
    6.     const n = s.length;
    7.     for (let i = 0; i < n; ++i) {
    8.         if (!isNaN(Number(s[i])) && s[i] !== ' ') {
    9.             num = num * 10 + s[i].charCodeAt() - '0'.charCodeAt();
    10.         }
    11.         if (isNaN(Number(s[i])) || i === n - 1) {
    12.             switch (preSign) {
    13.                 case '+':
    14.                     stack.push(num);
    15.                     break;
    16.                 case '-':
    17.                     stack.push(-num);
    18.                     break;
    19.                 case '*':
    20.                     stack.push(stack.pop() * num);
    21.                     break;
    22.                 default:
    23.                     stack.push(stack.pop() / num | 0);
    24.             }   
    25.             preSign = s[i];
    26.             num = 0;
    27.         }
    28.     }
    29.     let ans = 0;
    30.     while (stack.length) {
    31.         ans += stack.pop();
    32.     }
    33.     return ans;
    34. };
    36. 作者：LeetCode-Solution
    37. 链接：https://leetcode-cn.com/problems/basic-calculator-ii/solution/ji-ben-ji-suan-qi-ii-by-leetcode-solutio-cm28/
    38. 来源：力扣（LeetCode）
    39. 著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。