亚洲中字慕日产2020,大陆极品少妇内射AAAAAA,无码av大香线蕉伊人久久,久久精品国产亚洲av麻豆网站

資訊專欄INFORMATION COLUMN

[LeetCode] Validate IP Address

yeooo / 1445人閱讀

Problem

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).

However, we don"t replace a consecutive group of zero value with a single empty group using two consecutive colons (::) to pursue simplicity. For example, 2001:0db8:85a3::8A2E:0370:7334 is an invalid IPv6 address.

Besides, extra leading zeros in the IPv6 is also invalid. For example, the address 02001:0db8:85a3:0000:0000:8a2e:0370:7334 is invalid.

Note: You may assume there is no extra space or special characters in the input string.

Example 1:

Input: "172.16.254.1"

Output: "IPv4"

Explanation: This is a valid IPv4 address, return "IPv4".

Example 2:

Input: "2001:0db8:85a3:0:0:8A2E:0370:7334"

Output: "IPv6"

Explanation: This is a valid IPv6 address, return "IPv6".

Example 3:

Input: "256.256.256.256"

Output: "Neither"

Explanation: This is neither a IPv4 address nor a IPv6 address.

Solution

很爛，很無聊的題目

class Solution {
    public String validIPAddress(String IP) {
        if (validIPv4(IP)) return "IPv4";
        else if (validIPv6(IP)) return "IPv6";
        else return "Neither";
    }
    private boolean validIPv4(String IP) {
        if (IP == null || IP.length() < 7 || IP.charAt(0) == "." || IP.charAt(IP.length()-1) == ".") return false;
        String[] segments = IP.toLowerCase().split(".");
        if (segments.length != 4) return false;
        try {
            for (String segment: segments) {
                int value = Integer.parseInt(segment);
                if (value < 0 || value > 255 || 
                    (value != 0 && segment.startsWith("0")) || 
                    (value == 0 && segment.length() > 1)) {
                    return false;
                }
            }
        } catch (NumberFormatException e) {
            return false;
        }
        
        return true;
    }
    private boolean validIPv6(String IP) {
        if (IP == null || IP.length() < 15 || IP.charAt(0) == ":" || IP.charAt(IP.length()-1) == ":") return false;
        String[] segments = IP.toLowerCase().split(":");
        if (segments.length != 8) return false;
        for (String segment: segments) {
            System.out.println(segment);
            if (segment.length() == 0 || segment.length() > 4) return false;
            for (char ch: segment.toCharArray()) {
                if ((ch >= "a" && ch <= "f") || (ch >= "0" && ch <= "9")) continue;
                else return false;
            }
        }
        return true;
    }
}

文章版權(quán)歸作者所有，未經(jīng)允許請勿轉(zhuǎn)載,若此文章存在違規(guī)行為，您可以聯(lián)系管理員刪除。

轉(zhuǎn)載請注明本文地址：http://www.ezyhdfw.cn/yun/77141.html

相關(guān)文章

  • [Leetcode] Restore IP Address 修復(fù)IP地址

    摘要：四分法復(fù)雜度時間空間思路用三個點將字符串分成四段，驗證每一段是否是有效的。注意使用時要確保所得到數(shù)不會超界。代碼第一個分割點第二個分割點第三個分割點 Restore IP Address Given a string containing only digits, restore it by returning all possible valid IP address combin...

    VioletJack 評論0 收藏0

  • [Leetcode] Validate Binary Search Tree 驗證二叉搜索樹

    摘要：注意這里的結(jié)構(gòu)和不同的二叉樹遍歷一樣，如果到空節(jié)點就返回，否則遞歸遍歷左節(jié)點和右節(jié)點。唯一不同是加入了和，所以要在遞歸之前先判斷是否符合和的條件。代碼如果該節(jié)點大于上限返回假如果該節(jié)點小于下限返回假遞歸判斷左子樹和右子樹 Validate Binary Search Tree Given a binary tree, determine if it is a valid binary...

    fuchenxuan 評論0 收藏0

  • leetcode98. Validate Binary Search Tree

    摘要：題目要求檢驗二叉查找樹是否符合規(guī)則。二叉查找樹是指當(dāng)前節(jié)點左子樹上的值均比其小，右子樹上的值均比起大。因此在這里我們采用棧的方式實現(xiàn)中序遍歷，通過研究中序遍歷是否遞增來判斷二叉查找樹是否符合規(guī)則。 題目要求 Given a binary tree, determine if it is a valid binary search tree (BST). Assume a BST is...

    codercao 評論0 收藏0

  • leetcode98. Validate Binary Search Tree

    摘要：題目要求檢驗二叉查找樹是否符合規(guī)則。二叉查找樹是指當(dāng)前節(jié)點左子樹上的值均比其小，右子樹上的值均比起大。因此在這里我們采用棧的方式實現(xiàn)中序遍歷，通過研究中序遍歷是否遞增來判斷二叉查找樹是否符合規(guī)則。 題目要求 Given a binary tree, determine if it is a valid binary search tree (BST). Assume a BST is...

    AlphaWatch 評論0 收藏0

  • leetcode98. Validate Binary Search Tree

    摘要：題目要求判斷一個樹是否是二叉查找樹。二叉查找樹即滿足當(dāng)前節(jié)點左子樹的值均小于當(dāng)前節(jié)點的值，右子樹的值均大于當(dāng)前節(jié)點的值。思路一可以看到，對二叉查找樹的中序遍歷結(jié)果應(yīng)當(dāng)是一個遞增的數(shù)組。這里我們用堆棧的方式實現(xiàn)中序遍歷。 題目要求 given a binary tree, determine if it is a valid binary search tree (BST). Assu...

    songze 評論0 收藏0

發(fā)表評論

0條評論

yeooo

男|高級講師

TA的文章

閱讀更多
最新活動
閱讀需要支付1元查看
<