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很烂,很无聊的题目
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; } }
文章版权归作者所有,未经允许请勿转载,若此文章存在违规行为,您可以联系管理员删除。
转载请注明本文地址:https://www.ucloud.cn/yun/77141.html
摘要:四分法复杂度时间空间思路用三个点将字符串分成四段,验证每一段是否是有效的。注意使用时要确保所得到数不会超界。代码第一个分割点第二个分割点第三个分割点 Restore IP Address Given a string containing only digits, restore it by returning all possible valid IP address combin...
摘要:注意这里的结构和不同的二叉树遍历一样,如果到空节点就返回,否则递归遍历左节点和右节点。唯一不同是加入了和,所以要在递归之前先判断是否符合和的条件。代码如果该节点大于上限返回假如果该节点小于下限返回假递归判断左子树和右子树 Validate Binary Search Tree Given a binary tree, determine if it is a valid binary...
摘要:题目要求检验二叉查找树是否符合规则。二叉查找树是指当前节点左子树上的值均比其小,右子树上的值均比起大。因此在这里我们采用栈的方式实现中序遍历,通过研究中序遍历是否递增来判断二叉查找树是否符合规则。 题目要求 Given a binary tree, determine if it is a valid binary search tree (BST). Assume a BST is...
摘要:题目要求检验二叉查找树是否符合规则。二叉查找树是指当前节点左子树上的值均比其小,右子树上的值均比起大。因此在这里我们采用栈的方式实现中序遍历,通过研究中序遍历是否递增来判断二叉查找树是否符合规则。 题目要求 Given a binary tree, determine if it is a valid binary search tree (BST). Assume a BST is...
摘要:题目要求判断一个树是否是二叉查找树。二叉查找树即满足当前节点左子树的值均小于当前节点的值,右子树的值均大于当前节点的值。思路一可以看到,对二叉查找树的中序遍历结果应当是一个递增的数组。这里我们用堆栈的方式实现中序遍历。 题目要求 given a binary tree, determine if it is a valid binary search tree (BST). Assu...
阅读 1458·2021-09-02 13:57
阅读 1871·2019-08-30 15:55
阅读 2408·2019-08-30 15:54
阅读 2243·2019-08-30 15:44
阅读 2735·2019-08-30 13:18
阅读 484·2019-08-30 13:02
阅读 632·2019-08-29 18:46
阅读 1667·2019-08-29 11:25