golang判断ip地址是ipv4还是ipv6

leetcode连接

https://leetcode.cn/problems/validate-ip-address/description/

题目描述

func validIPAddress(queryIP string) string {
    if strings.Index(queryIP,".") >0 {
        // 验证是否是ipv4
        if isIpv4(queryIP){
            return "IPv4"
        }
        return "Neither"
    }


    if strings.Index(queryIP,":") > 0 {
        // 验证是否是ipv6
        if isIpv6(queryIP){
            return "IPv6"
        }
        return "Neither"
    }
    return "Neither"
}


func isIpv6(queryIP string) bool{
    var arr []string = strings.Split(queryIP,":")
    if len(arr)!=8 {
        return false
    }
    for i:=0;i<len(arr);i++ {
        if !ipv6Helper(arr[i]){
            return false
        }
    }
    return true
}


func ipv6Helper(ip string) bool{
    if len(ip)==0 || len(ip)>4{
        return false
    } 
    // 将16进制的ip转为int64看是否可以转 ，如果不能转，则返回err!=nil
    _,err := strconv.ParseInt(ip,16,64)
    return err == nil
}




func isIpv4(queryIP string) bool{
    var arr []string = strings.Split(queryIP,".")
    if len(arr)!=4 {
        return false
    }
    for i:=0;i<len(arr);i++  {
        if !ipv4Helper(arr[i]){
            return false
        }
    }
    return true
}


func ipv4Helper(ip string) bool{
    if len(ip)==0 || len(ip)>3 {
        return false
    }
    if ip[0]=='0' && len(ip)!=1{
        return false
    }
    // 将ip转为十进制的int
    i,err := strconv.Atoi(ip)
    if err != nil {
        return false
    }
    return i>=0 && i<=255
}

解题思路

通过left join + where is not null 过滤出在product表中的数据
通过group by + count(distinct(xxx)) 来过滤出所有商品

select c.customer_id
from customer c left join product p on c.product_key = p.product_key
where p.product_key is not null # 排除用户购买了一些不在product表中的商品
group by c.customer_id 
# 同一个用户可能买了一个商品多次，因此通过distinct去重
having count(distinct(p.product_key)) = (
    # 所有产品的数量
    select count(*)
    from product
)

题目描述

合作过至少三次的演员和导演

解题思路

按照演员id和导演id进行分组，多字段分组必须要所有字段一致才会分到同一组

select actor_id,director_id
from actordirector
group by actor_id,director_id
having count(*) >=3

题目描述

体育馆的人流量

解题思路

需要连续自增的三个id的时候，可以通过id = id-1，id=id-2进行leftjoin
union 具有去重功能， union all 不会进行去重
当我们查询的时候需要一些中间表的时候，这个中间表必须要起一个别名

select *
from(
select s1.*
from stadium s1 left join stadium s2 on s1.id = s2.id-1
left join stadium s3 on s1.id = s3.id-2 # 这个from语句体现了连续的id
where s2.id is not null and s3.id is not null and s1.people>=100 and s2.people>=100 and s3.people>=100
union 
select s2.*
from stadium s1 left join stadium s2 on s1.id = s2.id-1
left join stadium s3 on s1.id = s3.id-2
where s2.id is not null and s3.id is not null and s1.people>=100 and s2.people>=100 and s3.people>=100
union
select s3.*
from stadium s1 left join stadium s2 on s1.id = s2.id-1
left join stadium s3 on s1.id = s3.id-2
where s2.id is not null and s3.id is not null and s1.people>=100 and s2.people>=100 and s3.people>=100
) tmp order by id