題目
Given a valid (IPv4) IP address, return a defanged version of that IP address.
A defanged IP address replaces every period “.” with “[.]”.
Example 1:
Input: address = “1.1.1.1”
Output: “1[.]1[.]1[.]1”
Example 2:
Input: address = “255.100.50.0”
Output: “255[.]100[.]50[.]0”
Constraints:
The given address is a valid IPv4 address.
解析
顯然，就是將英文句號用括號包起來
答案
//將.換成[.]
return address.replace(".","[.]");
//將address按.拆分，然後分別加入[.]
return "[.]".join(address.split("."));
//基於正則表達式
return address.replaceAll("\\.", "[.]");
replace 與replaceAll
replace的參數是char和CharSequence，即可以支持字符的替換，也支持字符串的替換（CharSequence即字符串序列的意思,說白了也就是字符串）；
replaceAll的參數是regex，即基於規則表達式的替換，比如：可以通過replaceAll("\d", “*”)把一個字符串所有的數字字符都換成星號；
