一个行计算器,在测试系统检查时,给出错误:超出了程序执行时间限制(某些情况下程序进入死循环,我认为有些错误情况没有被考虑到。
使用两个堆栈实现,其中一个是操作,另一个是数字。
我从视频中获取了操作原理:https://youtu.be/Vk-tGND2bfc?si= c4RzRBDehfykDRlm
这是问题陈述:
实现一个函数
func Calc(expression string) (float64, error)
,其中表达式是由单字符标识符和算术运算符号组成的字符串表达式 输入数据 - 数字(有理数)、运算 +、-、*、/、优先运算(和 ) 写入错误的情况表达式,该函数会生成错误。
这是代码:
package main
import (
"errors"
"fmt"
"strconv"
"unicode"
)
func Calc(expression string) (float64, error) {
priority := map[rune]int{'+': 1, '-': 1, '*': 2, '/': 2}
var num []float64
var operator []rune
var hasNumber bool
for _, ch := range expression {
if unicode.IsDigit(ch) {
hasNumber = true
}
}
if !hasNumber {
return 0, errors.New("no number")
}
if len(expression) == 0 {
return 0, errors.New("empty expression")
}
check := rune(expression[len(expression)-1])
if !unicode.IsDigit(check) && check != ')' {
return 0, errors.New("invalid, last char is not digits or closing bracket")
}
applyOperator := func(a, b float64, op rune) float64 {
switch op {
case '+':
return a + b
case '-':
return a - b
case '*':
return a * b
case '/':
if b == 0 {
panic("division by zero")
}
return a / b
default:
return 0
}
}
calculate := func() {
if len(operator) == 0 || len(num) < 2 {
return
}
b := num[len(num)-1]
a := num[len(num)-2]
op := operator[len(operator)-1]
num = num[:len(num)-2]
operator = operator[:len(operator)-1]
result := applyOperator(a, b, op)
num = append(num, result)
}
for i := 0; i < len(expression); i++ {
ch := rune(expression[i])
if unicode.IsDigit(ch) || ch == '.' {
start := i
for i < len(expression) && (unicode.IsDigit(rune(expression[i])) || expression[i] == '.') {
i++
}
numer, err := strconv.ParseFloat(expression[start:i], 64)
if err != nil {
return 0, fmt.Errorf("failed to parse number: %v", err)
}
num = append(num, numer)
i--
} else if ch == '+' || ch == '-' {
if i == 0 || expression[i-1] == '(' || len(operator) > 0 && operator[len(operator)-1] == '(' {
num = append(num, 0)
}
for len(operator) > 0 && priority[operator[len(operator)-1]] >= priority[ch] {
calculate()
}
operator = append(operator, ch)
} else if ch == '*' || ch == '/' {
for len(operator) > 0 && priority[operator[len(operator)-1]] >= priority[ch] {
calculate()
}
operator = append(operator, ch)
} else if ch == '(' {
operator = append(operator, ch)
} else if ch == ')' {
for len(operator) > 0 && operator[len(operator)-1] != '(' {
calculate()
}
if len(operator) == 0 {
return 0, errors.New("mismatched parentheses")
}
operator = operator[:len(operator)-1]
} else if !unicode.IsSpace(ch) {
return 0, errors.New("invalid character")
}
}
for len(operator) > 0 {
calculate()
}
if len(num) == 1 {
return num[0], nil
}
return 0, errors.New("invalid expression")
}
1 个回答