Question

1108. Defanging an IP Address

中文文档

Description

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.

Solutions

Solution 1: Direct Replacement

We can directly replace the '.' in the string with '[.]'.

The time complexity is $O(n)$, where $n$ is the length of the string. Ignoring the space consumption of the answer, the space complexity is $O(1)$.

class Solution:
  def defangIPaddr(self, address: str) -> str:
    return address.replace('.', '[.]')

class Solution {
  public String defangIPaddr(String address) {
    return address.replace(".", "[.]");
  }
}

class Solution {
public:
  string defangIPaddr(string address) {
    for (int i = address.size(); i >= 0; --i) {
      if (address[i] == '.') {
        address.replace(i, 1, "[.]");
      }
    }
    return address;
  }
};

func defangIPaddr(address string) string {
  return strings.Replace(address, ".", "[.]", -1)
}

function defangIPaddr(address: string): string {
  return address.split('.').join('[.]');
}