Analog volume leetcode [normal] 093. Restore IP address

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093. Restore IP Address

It works IP Address It's just four integers （ Each integer is located in 0 To 255 Between the composition of , And it can't contain leading 0）, Use... Between integers ‘.’ Separate .

for example ：“0.1.2.201” and “192.168.1.1” yes It works IP Address , however “0.011.255.245”、“192.168.1.312” and “[email protected]” yes Invalid IP Address .
Given a string containing only numbers s , It is used to indicate a IP Address , Returns all possible valid IP Address , These addresses can be accessed through s Insert  ‘.’ To form . you You can't   Reorder or delete s Any number in . You can press whatever Return the answers in 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”]

Tips ：

1 <= s.length <= 20
s It's just numbers

source ： Power button （LeetCode）
link ：https://leetcode-cn.com/problems/restore-ip-addresses
Copyright belongs to the network . For commercial reprint, please contact the official authority , Non-commercial reprint please indicate the source .

Code ：

from leetcode_python.utils import *


def isLegal(ssss):
    if any(s.startswith('0') for s in ssss):return False
    return all(0<=int(s)<=255 for s in ssss)

class Solution:
    def restoreIpAddresses(self, s: str) -> List[str]:
        l = len(s)
        if l<4:return []
        res = []
        for id1 in range(1,l-2):
            for id2 in range(id1+1,l-1):
                for id3 in range(id2+1,l):
                    ssss = [s[:id1],s[id1:id2],s[id2:id3],s[id3:]]
                    if isLegal(ssss):
                        res.append('.'.join(*ssss))
        return res




def test(data_test):
    s = Solution()
    data = data_test  # normal
    # data = [List2Node(data_test[0])] # list turn node
    return s.getResult(*data)


def test_obj(data_test):
    result = [None]
    obj = Solution(*data_test[1][0])
    for fun, data in zip(data_test[0][1::], data_test[1][1::]):
        if data:
            res = obj.__getattribute__(fun)(*data)
        else:
            res = obj.__getattribute__(fun)()
        result.append(res)
    return result


if __name__ == '__main__':
    datas = [
        [],
    ]
    for data_test in datas:
        t0 = time.time()
        print('-' * 50)
        print('input:', data_test)
        print('output:', test(data_test))
        print(f'use time:{
      time.time() - t0}s')

remarks ：
GitHub：https://github.com/monijuan/leetcode_python

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leetcode_python.utils See the description on the summary page for details
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