深圳电子商城网站建设/佛山百度快照优化排名

题目描述

验证IP地址

思路

模拟

根据题意模拟即可。
观察到有效的IPv4地址中均存在’.‘符号，IPv4有4个部分用’.‘隔开。如果存在’.'，则考虑是否为IPv4地址，可以分割字符串，对于分割的每一部分，考虑是否满足以下条件，如果满足，则返回IPv4，如果不满足则返回neither：

• 长度是否在[1, 3]之间；
• 是否为纯数字；
• 值是否在[0, 255]之间；
• 是否不包含前导零。如果值为0，则该部分长度只能为1；如果值不为0，则该部分的第一个数字不能为0。

对于IPv6地址而言，均以’:‘分隔，共包含8个部分。如果存在’:'，则同样对地址进行分割，对每个部分进行判断，如果满足条件则返回IPv6，如果不满足则返回Neither。条件如下：

• 长度在[1, 4]之间；
• 是否只包含数字，或者a-f，或者A-F。

除了上述两种情况，如果无法找到对应数量的部分，给定的字符串也不是一个有效的IP地址。

Python实现

class Solution:def validIPAddress(self, queryIP: str) -> str:if queryIP.find('.') != -1:# 可能是IPv4last = -1for i in range(4):cur = (len(queryIP) if i == 3 else queryIP.find('.', last+1))if cur == -1:return "Neither"if not 1 <= cur-last-1 <= 3:return "Neither"addr = 0for j in range(last+1, cur):if not queryIP[j].isdigit():return "Neither"addr = addr * 10 + int(queryIP[j])if addr > 255 or (addr > 0 and queryIP[last+1] == '0') or (addr == 0 and cur-last-1>1):return "Neither"last = curreturn "IPv4"else:# 可能是IPv6last = -1for i in range(8):cur = (len(queryIP) if i == 7 else queryIP.find(":", last+1))if cur == -1:return "Neither"if not 1 <= cur-last-1 <= 4:return "Neither"for j in range(last+1, cur):if not queryIP[j].isdigit() and not ("a" <= queryIP[j].lower() <= "f"):return "Neither"last = curreturn "IPv6"

Java实现

class Solution {public String validIPAddress(String queryIP) {if (queryIP.indexOf('.') >= 0) {// IPv4int last = -1;for (int i = 0; i < 4; ++i) {int cur = (i == 3 ? queryIP.length() : queryIP.indexOf('.', last + 1));if (cur < 0) {return "Neither";}if (cur - last - 1 < 1 || cur - last - 1 > 3) {return "Neither";}int addr = 0;for (int j = last + 1; j < cur; ++j) {if (!Character.isDigit(queryIP.charAt(j))) {return "Neither";}addr = addr * 10 + (queryIP.charAt(j) - '0');}if (addr > 255) {return "Neither";}if (addr > 0 && queryIP.charAt(last + 1) == '0') {return "Neither";}if (addr == 0 && cur - last - 1 > 1) {return "Neither";}last = cur;}return "IPv4";} else {// IPv6int last = -1;for (int i = 0; i < 8; ++i) {int cur = (i == 7 ? queryIP.length() : queryIP.indexOf(':', last + 1));if (cur < 0) {return "Neither";}if (cur - last - 1 < 1 || cur - last - 1 > 4) {return "Neither";}for (int j = last + 1; j < cur; ++j) {if (!Character.isDigit(queryIP.charAt(j)) && !('a' <= Character.toLowerCase(queryIP.charAt(j)) && Character.toLowerCase(queryIP.charAt(j)) <= 'f')) {return "Neither";}}last = cur;}return "IPv6";}}
}

C++实现

class Solution {
public:string validIPAddress(string queryIP) {if (queryIP.find('.') != string::npos) {// IPv4int last = -1;for (int i = 0; i < 4; ++i) {int cur = (i == 3 ? queryIP.size() : queryIP.find('.', last + 1));if (cur == string::npos) {return "Neither";}if (cur - last - 1 < 1 || cur - last - 1 > 3) {return "Neither";}int addr = 0;for (int j = last + 1; j < cur; ++j) {if (!isdigit(queryIP[j])) {return "Neither";}addr = addr * 10 + (queryIP[j] - '0');}if (addr > 255) {return "Neither";}if (addr > 0 && queryIP[last + 1] == '0') {return "Neither";}if (addr == 0 && cur - last - 1 > 1) {return "Neither";}last = cur;}return "IPv4";}else {// IPv6int last = -1;for (int i = 0; i < 8; ++i) {int cur = (i == 7 ? queryIP.size() : queryIP.find(':', last + 1));if (cur == string::npos) {return "Neither";}if (cur - last - 1 < 1 || cur - last - 1 > 4) {return "Neither";}for (int j = last + 1; j < cur; ++j) {if (!isdigit(queryIP[j]) && !('a' <= tolower(queryIP[j]) && tolower(queryIP[j]) <= 'f')) {return "Neither";}}last = cur;}return "IPv6";}}
};