本题也不难，最开始我用deque的方法做出了，但是更加费空间，所以看了下答案，答案思路一致，但是简化的更好：

    • 遇到*和/的时侯，进行计算

    • 遇到+和-，推进Stack<Integer>中

      • 根据之前的operator，来决定存正还是负

    • 空间复杂度: 227. Basic Calculator II - 图1

    • 时间复杂度:227. Basic Calculator II - 图2

    代码远比描述更好懂，直接上代码了：

    1. class Solution {
    2.   public int calculate(String s) {
    3.     int result = 0;
    4.     Stack<Integer> stack = new Stack<>();
    6.     char operator = '+';
    7.     int curr = 0;
    8.     for (int i = 0; i < s.length(); ++i) {
    9.       char ch = s.charAt(i);
    11.       if (ch == ' ') {
    12.         continue;
    13.       }
    14.       else if (ch >= '0' && ch <= '9') {
    15.         curr = curr * 10 + (ch - '0');
    16.       }
    17.       else {
    18.         if (operator == '+') {
    19.           stack.push(curr);
    20.         }
    21.         else if (operator == '-') {
    22.           stack.push(-curr);
    23.         }
    24.         else if (operator == '*') {
    25.           stack.push(stack.pop() * curr);
    26.         }
    27.         else if (operator == '/') {
    28.           stack.push(stack.pop() / curr);
    29.         }
    30.         curr = 0;
    31.         operator = ch;
    32.       }
    33.     }
    35.     if (operator == '+') {
    36.       stack.push(curr);
    37.     }
    38.     else if (operator == '-') {
    39.       stack.push(-curr);
    40.     }
    41.     else if (operator == '*') {
    42.       stack.push(stack.pop() * curr);
    43.     }
    44.     else if (operator == '/') {
    45.       stack.push(stack.pop() / curr);
    46.     }
    48.     while (!stack.isEmpty()) {
    49.       result += stack.pop();
    50.     }
    52.     return r