Basic Calculator II

Description

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.

Code

和 Basic Calculator 一樣的解法。

Time Complexity: $O(n)$, Space Complexity: $O(n)$

class Result {
public:
    int equationResult;
    int charsInEquation;
    Result(int equationResult, int charsInEquation) {
        this->equationResult = equationResult;
        this->charsInEquation = charsInEquation;
    }
};
 
class Solution {
public:
    int calculate(string s) {
        return calculateResult(s).equationResult;
    }
 
    void update(stack<int>& st, int num, int sign) {
        if(sign == '+') st.push(num);
        if(sign == '-') st.push(-num);
        if(sign == '*') {
            int num2 = st.top();
            st.pop();  
            st.push(num2 * num);  
        }
        if(sign == '/') {
            int num2 = st.top();
            st.pop();  
            st.push(num2 / num);  
        }
    }
 
    int sum(stack<int> st) {
        int result = 0;
        while(!st.empty()){
            result += st.top();
            st.pop();
        }
        return result;
    }
 
    Result calculateResult(string s) {
        stack<int> st;
        int i = 0, num = 0;
        char sign = '+';
        unordered_set<char> ops = {'+', '-', '/', '*'};
 
        while(i < s.length()) {
            char c = s[i];
            if(isdigit(c)) {
                num = num * 10 + (c - '0');
            } else if(ops.find(c) != ops.end()) {
                update(st, num, sign);
                num = 0;
                sign = c;
            } else if(c == '(') {
                Result innerEquationResult = calculateResult(s.substr(i+1));
                num = innerEquationResult.equationResult;
                i += innerEquationResult.charsInEquation;
            } else if(c == ')') {
                update(st, num, sign);
                return Result(sum(st), i+1); 
            }
            i++;
        }
 
        update(st, num, sign);
        return Result(sum(st), i);
    }
};

Source

• Basic Calculator II - LeetCode