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Leetcode - Validate IP Address

violet posted @ Jun 17, 2020 01:17:29 AM in 算法 with tags Algorithm Golang string , 395 阅读

https://leetcode.com/problems/validate-ip-address/

Write a function to check whether an input string is a valid IPv4 address or IPv6 address or neither.

IPv4 addresses are canonically represented in dot-decimal notation, which consists of four decimal numbers, each ranging from 0 to 255, separated by dots ("."), e.g.,172.16.254.1;

Besides, leading zeros in the IPv4 is invalid. For example, the address 172.16.254.01 is invalid.

IPv6 addresses are represented as eight groups of four hexadecimal digits, each group representing 16 bits. The groups are separated by colons (":"). For example, the address 2001:0db8:85a3:0000:0000:8a2e:0370:7334 is a valid one. Also, we could omit some leading zeros among four hexadecimal digits and some low-case characters in the address to upper-case ones, so 2001:db8:85a3:0:0:8A2E:0370:7334 is also a valid IPv6 address(Omit leading zeros and using upper cases).

 

func validIPAddress(IP string) string {
    resultV4 := check4(strings.Split(IP, "."))
    resultV6 := check6(strings.Split(IP, ":"))
    if resultV4 {
        return "IPv4"
    }
    if resultV6 {
        return "IPv6"
    }
    return "Neither"
}

func check4(strs []string) bool {
    if len(strs) != 4 {
        return false
    }
    for _, str := range strs {
        val, err := strconv.Atoi(str)
        if err != nil {
            return false
        }
        if val > 255 || val < 0 {
            return false
        }
        if str[0] == '0' && len(str) > 1 {
            return false
        }
        if str[0] == '-' || str[0] == '+' {
            return false
        }
    }
    return true
}

func check6(strs []string) bool {
    if len(strs) != 8 {
        return false
    }
    for _, str := range strs {
        if len(str) > 4 || len(str) == 0 {
            return false
        }
        for i := 0; i < len(str); i++ {
            if !((str[i] >= '0' && str[i] <= '9') || (str[i] >= 'a' && str[i] <= 'f') || (str[i] >= 'A' && str[i] <= 'F')) {
                return false
            }
        }
    }
    return true
}
+1 Question Paper 20 说:
Feb 16, 2023 08:06:11 AM

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