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Formatted question description: https://leetcode.ca/all/1108.html

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 address is a valid IPv4 address.

Related Topics:
String

Solution 1.

• Java
• C++
• Python
• Go
• TypeScript

• class Solution {
      public String defangIPaddr(String address) {
          address = address.replaceAll("\\.", "[\\.]");
          return address;
      }
  }
  
  ############
  
  class Solution {
      public String defangIPaddr(String address) {
          return address.replace(".", "[.]");
      }
  }
  

• // OJ: https://leetcode.com/problems/defanging-an-ip-address/
  // Time: O(N)
  // Space: O(1)
  class Solution {
  public:
      string defangIPaddr(string address) {
          string ans;
          for (char c : address) {
              if (c == '.') ans += "[.]";
              else ans += c;
          }
          return ans;
      }
  };
  

• class Solution:
      def defangIPaddr(self, address: str) -> str:
          return address.replace('.', '[.]')
  
  ############
  
  # 1108. Defanging an IP Address
  # https://leetcode.com/problems/defanging-an-ip-address/
  
  class Solution:
      def defangIPaddr(self, address: str) -> str:
          address = address.replace('.', '[.]')
          
          return address
  
  

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

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

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