Restore ip addresses
class Solution:
def __init__(self):
self.res = []
self.path = []
def backtrack(self, s, startIndex):
if len(self.path) > 4:
return None
if startIndex == len(s) and len(self.path) == 4:
self.res.append(".".join(self.path))
return None
for i in range(startIndex, len(s)):
if self.check(s[startIndex:i + 1]):
self.path.append(s[startIndex:i + 1])
self.backtrack(s, i + 1)
self.path.pop()
def check(self, s):
if len(s) > 1 and s[0] == '0':
return False
if 0 <= int(s) <= 255:
return True
else:
return False
def restoreIpAddresses(self, s: str) -> List[str]:
startIndex = 0
self.backtrack(s, startIndex)
return self.res
Restore IP Addresses
A valid IP address consists of exactly four integers separated by single dots. Each integer is between 0 and 255 (inclusive) and cannot have leading zeros.
- For example,
"0.1.2.201"and"192.168.1.1"are valid IP addresses, but"0.011.255.245","192.168.1.312"and"192.168@1.1"are invalid IP addresses.
Given a string s containing only digits, return all possible valid IP addresses that can be formed by inserting dots into s. You are not allowed to reorder or remove any digits in s. You may return the valid IP addresses in any order.
Example 1:
Input: s = "25525511135" Output: ["255.255.11.135","255.255.111.35"]
Example 2:
Input: s = "0000" Output: ["0.0.0.0"]
Example 3:
Input: s = "101023" Output: ["1.0.10.23","1.0.102.3","10.1.0.23","10.10.2.3","101.0.2.3"]
Constraints:
1 <= s.length <= 20sconsists of digits only.