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basic-calculator

224. Basic Calculator

Given a string s representing a valid expression, implement a basic calculator to evaluate it, and return the result of the evaluation.

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 = " 2-1 + 2 " Output: 3

Example 3:

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

Constraints:

  • 1 <= s.length <= 3 * 105
  • s consists of digits, '+', '-', '(', ')', and ' '.
  • s represents a valid expression.
  • '+' is not used as a unary operation (i.e., "+1" and "+(2 + 3)" is invalid).
  • '-' could be used as a unary operation (i.e., "-1" and "-(2 + 3)" is valid).
  • There will be no two consecutive operators in the input.
  • Every number and running calculation will fit in a signed 32-bit integer.
class Solution {
public:
int calculate(string s) {
stack<int> stk;
int res = 0;
int sign = 1;
int operand = 0;
for(int i = 0; i < s.size(); i++) {
char c = s[i];
if(isdigit(c)) {
operand = 10 * operand + (c - '0');
} else if(c == '+') {
res += sign * operand;
sign = 1;
// 已經將答案存到 res 了,operand reset 為 0
operand = 0;
} else if(c == '-') {
res += sign * operand;
sign = -1;
// 已經將答案存到 res 了,operand reset 為 0
operand = 0;
} else if(c == '(') {
// 將結果暫時存到 stack
stk.push(res);

// 也將 sign 暫時存到 stack
stk.push(sign);

// reset 結果為 0,重新計算
res = 0;

// reset sign 為 1,重新計算
sign = 1;
} else if(c == ')') {
// 加上目前的計算
res += sign * operand;

// 乘上之前運算的結果的 sign
res *= stk.top(); stk.pop();

// 加上之前運算的結果
res += stk.top(); stk.pop();

// reset operand 為 0
operand = 0;
}
}
return res + (sign * operand);
}
};
  • T: O(N)O(N)
  • S: O(N)O(N)