Memory Usage:14 MB, less than5.10% of Python3 online submissions for Defanging an IP Address. 【解法】 简洁,一句话概括上面的两种解法。 defdefangIPaddr(self, address: str) ->str:returnaddress.replace('.','[.]')defdefangIPaddr(self, address: str) ->str:return'[.]'.join(address.split...
Input:address = "255.100.50.0"Output:"255[.]100[.]50[.]0" Constraints: The givenaddressis a valid IPv4 address. Solution classSolution {publicString defangIPaddr(String address) { String result= "";if(address==null||address.length()==0){returnresult; }for(inti =0; i<address.length()...
publicStringdefangIPaddr(String address){returnaddress.replace(".","[.]"); } 03 第二种解法 不使用String的replace方法,使用StringBuilder,依次遍历address中的字符,碰到"."就拼接"[.]",其他字符不变,最后返回StringBuilder对象转成的String。 publicStringdefangIPaddr2(String address){StringBuildersb=newStringB...
classSolution {publicString defangIPaddr(String address) {char[] charsArr =address.toCharArray(); StringBuilder sb=newStringBuilder();for(charc : charsArr) {if(c == '.') sb.append("[.]");elsesb.append(c); }returnsb.toString(); } } 参考资料:n/a LeetCode 题目列表 -LeetCode Question...
The givenaddressis a valid IPv4 address. Solution1: class Solution: def defangIPaddr(self, address: str) -> str: output = "" for i in range(len(address)): if address[i] == ".": output += "[.]" else: output += address[i] ...