LeetCode-in-Go
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
type Solution struct {
i int
}
func calculate(s string) int {
sol := &Solution{}
return sol.helper([]rune(s))
}
func (sol *Solution) helper(ca []rune) int {
num := 0
prenum := 0
isPlus := true
for ; sol.i < len(ca); sol.i++ {
c := ca[sol.i]
if c != ' ' {
if c >= '0' && c <= '9' {
if num == 0 {
num = int(c - '0')
} else {
num = num*10 + int(c-'0')
}
} else if c == '+' {
prenum += num * sol.boolToInt(isPlus)
isPlus = true
num = 0
} else if c == '-' {
prenum += num * sol.boolToInt(isPlus)
num = 0
isPlus = false
} else if c == '(' {
sol.i++
prenum += sol.helper(ca) * sol.boolToInt(isPlus)
isPlus = true
num = 0
} else if c == ')' {
return prenum + num*sol.boolToInt(isPlus)
}
}
}
return prenum + num*sol.boolToInt(isPlus)
}
func (sol *Solution) boolToInt(b bool) int {
if b {
return 1
}
return -1
}