讀處

【原題目】:

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.

參考 : https://leetcode.com/problems/defanging-an-ip-address/

【My Solution】:

本來想跑迴圈，然後進行判斷後做替換，但後來發現Java String有replace的寫法，看起來比較乾淨俐落

public String replace(char oldChar,
                      char newChar)

replace()方法透過用newChar取代字串中所有的oldchar