8. 字符串转换整数 (atoi)

算法原链接

https://leetcode-cn.com/problems/string-to-integer-atoi/

解法

package main
import (
    "fmt"
    "math"
    "regexp"
    "strconv"
    "strings"
)
func main() {
    str := "9223372036854775808"
    //ii := 922337203685477580
    //fmt.Println(ii * 10 + 8)
    //i := myAtoi(str)
    //fmt.Println(i)
    i := myAtoi1(str)
    fmt.Println(i)
}
// 自己的解法，耗时很短，但是相对占内存一些
func myAtoi(s string) int {
    reg := regexp.MustCompile(`^\s*[+-]?\d*`)
    result := reg.FindString(s)
    result = strings.Trim(result, " ")
    if len(result) == 0 {
        return 0
    }
    num, _ := strconv.Atoi(result)
    if num > math.MaxInt32 {
        return math.MaxInt32
    }
    if num < math.MinInt32 {
        return math.MinInt32
    }
    return num
}
// 这个解法参考了官方的自动机解法思路
// 纯操作数字会导致数字溢出变成负数，所以用字符串拼接的方式再转数字
func myAtoi1(s string) int {
    var i int
    //var flag = 1
    // start signed in_number end
    state := "start"
    str := ""
    for _, ch1 := range s {
        if state == "start" && ch1 == 32 {
            continue
        } else if state == "start" && (ch1 == 45 || ch1 == 43) {
            state = "signed"
            if ch1 == 45 {
                str = "-"
            }
            //else {
            //    flag = 1
            //}
        } else if state != "end" && ch1 >= 48 && ch1 <= 57 {
            state = "in_number"
            //c1, _ := strconv.Atoi(string(ch1))
            //i = i*10 + c1
            str += string(ch1)
        } else {
            state = "end"
            break
        }
    }
    //i *= flag
    i, _ = strconv.Atoi(str)
    if i > math.MaxInt32 {
        return math.MaxInt32
    }
    if i < math.MinInt32 {
        return math.MinInt32
    }
    return i
}