227. Basic Calculator II

[fockspaces]

Given a string s which represents an expression, evaluate this expression and return its value.

The integer division should truncate toward zero.

You may assume that the given expression is always valid. All intermediate results will be in the range of [-231, 231 - 1].

Note: You are not allowed to use any built-in function which evaluates strings as mathematical expressions, such as eval().

Example 1:

Input: s = "3+2*2" Output: 7 Example 2:

Input: s = " 3/2 " Output: 1 Example 3:

Input: s = " 3+5 / 2 " Output: 5

Constraints:

1 <= s.length <= 3 105 s consists of integers and operators ('+', '-', '', '/') separated by some number of spaces. s represents a valid expression. All the integers in the expression are non-negative integers in the range [0, 231 - 1]. The answer is guaranteed to fit in a 32-bit integer.

[fockspaces]

解法：自加 num，遇到正負 push +- num，遇到乘除拿前一個跟當前 num 做 基本上我之後也寫不出這解法，實在太巧妙，可能要用到再特別記吧

class Solution {
public:
    int calculate(string s) {
        s += '+'; int num = 0; char op = '+';
        stack<int> nums;
        for(char c : s) {
            if(isdigit(c)) num = num * 10 + (c - '0');
            else if(c != ' ') {
                if(op == '+') nums.push(num);
                else if(op == '-') nums.push(-num);
                else if(op == '*') {
                    int prev_num = nums.top(); nums.pop();
                    nums.push(prev_num * num); 
                } else {
                    int prev_num = nums.top(); nums.pop();
                    nums.push(prev_num / num);
                }
                num = 0; op = c;
            }
        }
        while(nums.size()) {num += nums.top(); nums.pop();}
        return num;
    }
};