【LeetCode 224】Basic Calculator

題目描述

Implement a basic calculator to evaluate a simple expression string.

The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and empty spaces .

Example 1:

Input: "1 + 1"
Output: 2

Example 2:

Input: " 2-1 + 2 "
Output: 3

Example 3:

Input: "(1+(4+5+2)-3)+(6+8)"
Output: 23

Note:

You may assume that the given expression is always valid.
Do not use the eval built-in library function.

思路

符號標記正負，用一個數字記錄當前的結果，遇見左括號時，入棧，先計算括號部分，遇見右括號時，出棧，計算之前的和括號中的結果。

代碼

class Solution {
public:
    int calculate(string s) {
        int sign = 1, res = 0;
        int n = s.length();
        stack<int> st;
        
        for (int i=0; i<n; ++i) {
            if (s[i] >= '0') {
                int num = 0;
                while(s[i] >= '0') num = num * 10 + (s[i++] - '0');
                i--;
                num *= sign;
                res += num;
            }else if (s[i] == '+') {
                sign = 1;
            }else if (s[i] == '-') {
                sign = -1;
            }else if (s[i] == '(') {
                st.push(res);
                st.push(sign);
                res = 0;
                sign = 1;
            }else if (s[i] == ')') {
                res *= st.top(); st.pop();
                res += st.top(); st.pop();
            }
        }
        
        return res;
    }
};