Restore IP Addresses Problem
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:
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Input: s = "25525511135"
Output: ["255.255.11.135","255.255.111.35"]
Example 2:
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Input: s = "0000"
Output: ["0.0.0.0"]
Example 3:
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Input: s = "1111"
Output: ["1.1.1.1"]
Example 4:
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Input: s = "010010"
Output: ["0.10.0.10","0.100.1.0"]
Example 5:
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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
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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[0]) + "." +
to_string(address[1]) + "." + to_string(address[2]) +
"." + to_string(address[3]);
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) {
address.push_back(num);
backtrack(address, i+1);
address.pop_back();
} else {
break;
}
}
}
vector<string> restoreIpAddresses(string s) {
origS = s;
vector<int> address;
backtrack(address, 0);
return ans;
}
};