給一組有效的 IPv4 的 IP 地址，返回該 IP 地址的默認版本。
然後這個違規的 IP 地址會將句號 "." 用 "[.]" 取代。
LeetCode eetCode 題目連結
https://leetcode.com/problems/defanging-an-ip-address/
題目
1108. Defanging an IP Address
Easy
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
addressis a valid IPv4 address.
Accept 作法
Runtime: 188 ms
Memory: 36.6 mb
class Solution {
fun defangIPaddr(address: String): String {
val result = address.replace(".","[.]")
return result
}
}
後來選擇還是不要套現成的 API 好了。就有第二種作法。
Runtime: 140 ms
Memory: 31.6 mb
Kotlin
class Solution {
fun defangIPaddr(address: String): String {
var newAddress = "";
for(i in 0..address.length-1){
if(address.get(i).toString().equals(".")){
newAddress = newAddress+"["+ address.get(i)+"]"
}else{
newAddress = newAddress+address.get(i)
}
}
return newAddress
}
}
這邊 Kotlin 跟 Java 的 String 不一樣的是它拿每一個字元直接用 get 不是 charAt ， get 回來轉型成 String 才能用 equals 來比對。
還有第三種作法，String 最常被討論如果有大量字串處理時有效能問題，用 StringBuilder 比較好，於是這題就拿來實驗看看，但結果時間一樣，我想是因為本身字串長度就滿短的，所以沒差。
Runtime: 140 ms
Memory: 31.6 mb
Kotlin
class Solution {
fun defangIPaddr(address: String): String {
var newAddress = StringBuilder();
for(i in 0..address.length-1){
if(address.get(i).toString().equals(".")){
newAddress.append("["+ address.get(i)+"]")
}else{
newAddress.append(address.get(i))
}
}
return newAddress.toString()
}
}
更多 LeetCode 相關資源
複習程式面試書籍
除了 LeetCode 練習外，我也入手了這本，題庫來自真正的面試，並非摘自教科書。它們反映出頂尖公司真正會出的題目，你可以藉此做好充分準備。
需要的話可以看看，寫得很仔細。
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