[LC] 224. Basic Calculator
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 .
Example 1:
Input: "1 + 1"
Output: 2
Example 2:
Input: " 2-1 + 2 "
Output: 3
Example 3:
Input: "(1+(4+5+2)-3)+(6+8)"
Output: 23
Note:
- You may assume that the given expression is always valid.
- Do not use the
eval
built-in library function.
class Solution(object):
def calculate(self, s):
"""
:type s: str
:rtype: int
"""
sign = 1
res = 0
stack = []
index = 0
while index < len(s):
char = s[index]
if char.isdigit():
num = int(char)
while index + 1 < len(s) and s[index + 1].isdigit():
num = 10 * num + int(s[index + 1])
index += 1
res += sign * num
elif char == '+':
sign = 1
elif char == '-':
sign = -1
elif char == '(':
stack.append(res)
stack.append(sign)
res = 0
sign = 1
elif char == ')':
res = stack.pop() * res + stack.pop()
index += 1
return res
class Solution {
public int calculate(String s) {
char[] charArr = s.toCharArray();
LinkedList<Integer> stack = new LinkedList<>();
int sign = 1;
int num = 0;
for (int i = 0; i < charArr.length; i++) {
char cur = charArr[i];
if (Character.isDigit(cur)) {
int count = cur - '0';
while (i + 1 < charArr.length && Character.isDigit(charArr[i + 1])) {
// need to use charArr[i + 1]
count = 10 * count + charArr[i + 1] - '0';
i += 1;
}
num = num + count * sign;
} else if (cur == '+') {
sign = 1;
} else if (cur == '-') {
sign = -1;
} else if (cur == '(') {
stack.offerFirst(num);
stack.offerFirst(sign);
num = 0;
sign = 1;
} else if (cur == ')') {
num = num * stack.pollFirst() + stack.pollFirst();
}
}
return num;
}
}
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