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772. Basic Calculator III (Hard)

Implement a basic calculator to evaluate a simple expression string.

The expression string contains only non-negative integers, '+', '-', '*', '/' operators, and open '(' and closing parentheses ')'. 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 = "1+1"
Output: 2

Example 2:

Input: s = "6-4/2"
Output: 4

Example 3:

Input: s = "2*(5+5*2)/3+(6/2+8)"
Output: 21

Example 4:

Input: s = "(2+6*3+5-(3*14/7+2)*5)+3"
Output: -12

Example 5:

Input: s = "0"
Output: 0

 

Constraints:

• 1 <= s <= 104
• s consists of digits, '+', '-', '*', '/', '(', and ')'.
• s is a valid expression.

Companies:
Amazon, Facebook, Microsoft, Uber

Related Topics:
Math, String, Stack, Recursion

Similar Questions:

• Basic Calculator (Hard)
• Basic Calculator II (Medium)
• Basic Calculator IV (Hard)
• Build Binary Expression Tree From Infix Expression (Hard)

Solution 1.

// OJ: https://leetcode.com/problems/basic-calculator-iii/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
    stack<long> num;
    stack<char> op;
    unordered_map<char, int> priority{{'+', 1}, {'-', 1}, {'*', 2}, {'/', 2}};
    void eval() {
        long b = num.top(); num.pop();
        char c = op.top(); op.pop();
        switch (c) {
            case '+': num.top() += b; break;
            case '-': num.top() -= b; break;
            case '*': num.top() *= b; break;
            case '/': num.top() /= b; break;
        }
    }
public:
    int calculate(string s) {
        for (int i = 0, N = s.size(); i < N; ++i) {
            if (s[i] == ' ') continue;
            if (isdigit(s[i])) {
                long n = 0;
                while (i < N && isdigit(s[i])) n = n * 10 + s[i++] - '0';
                --i;
                num.push(n);
            } else if (s[i] == '(') op.push(s[i]);
            else if (s[i] == ')') {
                while (op.top() != '(') eval();
                op.pop();
            } else {
                while (op.size() && op.top() != '(' && priority[op.top()] >= priority[s[i]]) eval();
                op.push(s[i]);
            }
        }
        while (op.size()) eval();
        return num.top();
    }
};