# 772. Basic Calculator III

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 ``.

The expression string contains only non-negative integers, `+``-``*``/` operators , open `(` and closing parentheses `)` and empty spaces ``. 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 `[-2147483648, 2147483647]`.

Some examples:

``````"1 + 1" = 2
" 6-4 / 2 " = 4
"2*(5+5*2)/3+(6/2+8)" = 21
"(2+6* 3+5- (3*14/7+2)*5)+3"=-12
``````

Note: Do not use the `eval` built-in library function.

``````class Solution {
public:
int calculate(string s) {
int n = s.size(), num = 0, curRes = 0, res = 0;
char op = '+';
for (int i = 0; i < n; ++i) {
char c = s[i];
if (c >= '0' && c <= '9') {
num = num * 10 + c - '0';
} else if (c == '(') {
int j = i, cnt = 0;
for (; i < n; ++i) {
if (s[i] == '(') ++cnt;
if (s[i] == ')') --cnt;
if (cnt == 0) break;
}
num = calculate(s.substr(j + 1, i - j - 1));
}
if (c == '+' || c == '-' || c == '*' || c == '/' || i == n - 1) {
switch (op) {
case '+': curRes += num; break;
case '-': curRes -= num; break;
case '*': curRes *= num; break;
case '/': curRes /= num; break;
}
if (c == '+' || c == '-' || i == n - 1) {
res += curRes;
curRes = 0;
}
op = c;
num = 0;
}
}
return res;
}
};
``````

Github 同步地址：

https://github.com/grandyang/leetcode/issues/772

Basic Calculator IV

Basic Calculator II

Basic Calculator

https://leetcode.com/problems/basic-calculator-iii/

https://leetcode.com/problems/basic-calculator-iii/discuss/113597/C++-recursive

https://leetcode.com/problems/basic-calculator-iii/discuss/113593/C++-Consise-Solution

https://leetcode.com/problems/basic-calculator-iii/discuss/113592/Development-of-a-generic-solution-for-the-series-of-the-calculator-problems

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