LeetCode in TypeScript
224. Basic Calculator
Hard
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 * 105sconsists of digits,'+','-','(',')', and' '.srepresents 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.
Solution
function calculate(s: string): number {
let i = 0
function helper(ca: string[]): number {
let num = 0
let prenum = 0
let isPlus = true
while (i < ca.length) {
const c = ca[i]
if (c !== ' ') {
if (c >= '0' && c <= '9') {
num = num * 10 + parseInt(c)
} else if (c === '+') {
prenum += num * (isPlus ? 1 : -1)
isPlus = true
num = 0
} else if (c === '-') {
prenum += num * (isPlus ? 1 : -1)
isPlus = false
num = 0
} else if (c === '(') {
i++
num = helper(ca)
} else if (c === ')') {
prenum += num * (isPlus ? 1 : -1)
return prenum
}
}
i++
}
return prenum + num * (isPlus ? 1 : -1)
}
return helper(s.split(''))
}
export { calculate }