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);
}
};
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