## Description

LeetCode Problem 93.

Given a string s containing only digits, return all possible valid IP addresses that can be obtained from s. You can return them in any order.

A valid IP address consists of exactly four integers, each integer is between 0 and 255, separated by single dots and cannot have leading zeros. For example, “0.1.2.201” and “192.168.1.1” are valid IP addresses and “0.011.255.245”, “192.168.1.312” and “192.168@1.1” are invalid IP addresses.

Example 1:

``````1
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Input: s = "25525511135"
Output: ["255.255.11.135","255.255.111.35"]
``````

Example 2:

``````1
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Input: s = "0000"
Output: ["0.0.0.0"]
``````

Example 3:

``````1
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Input: s = "1111"
Output: ["1.1.1.1"]
``````

Example 4:

``````1
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Input: s = "010010"
Output: ["0.10.0.10","0.100.1.0"]
``````

Example 5:

``````1
2
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:

• 0 <= s.length <= 20
• s consists of digits only.

## Sample C++ Code

``````1
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class Solution {
public:
vector<string> ans;
string origS;

void backtrack(vector<int>& address, int idx) {
if (address.size() == 4 && idx == origS.size()) {
string S = to_string(address) + "." +
ans.push_back(S);
}
if (address.size() >= 4 || idx >= origS.size())
return;

int num = 0;
for (int i = idx; i < origS.size(); i ++) {
num = num * 10 + origS[i] - '0';
if (origS[idx] == '0' && i > idx)
break;
if (num <= 255) {
} else {
break;
}
}
}

origS = s;
return ans;
}
};
``````