Given a string containing only digits, restore it by returning all possible valid IP address combinations.
Have you met this question in a real interview?
Yes
Example
Given "25525511135", return
[
"255.255.11.135",
"255.255.111.35"
]
Order does not matter.
LeetCode上的原题，请参见我之前的博客Restore IP Addresses。
解法一：
class Solution {
public:
/**
* @param s the IP string
* @return All possible valid IP addresses
*/
vector<string> restoreIpAddresses(string& s) {
vector<string> res;
helper(s, , "", res);
return res;
}
void helper(string s, int k, string out, vector<string>& res) {
if (k == ) {
if (s.empty()) res.push_back(out);
return;
}
for (int i = ; i < ; ++i) {
if (s.size() < i) break;
int val = stoi(s.substr(, i));
if (val > || i != to_string(val).size()) continue;
helper(s.substr(i), k - , out + s.substr(, i) + (k == ? "" : "."), res);
}
}
};
解法二：
class Solution {
public:
/**
* @param s the IP string
* @return All possible valid IP addresses
*/
vector<string> restoreIpAddresses(string& s) {
vector<string> res;
for (int a = ; a < ; ++a)
for (int b = ; b < ; ++b)
for (int c = ; c < ; ++c)
for (int d = ; d < ; ++d)
if (a + b + c + d == s.size()) {
int A = stoi(s.substr(, a));
int B = stoi(s.substr(a, b));
int C = stoi(s.substr(a + b, c));
int D = stoi(s.substr(a + b + c, d));
if (A <= && B <= && C <= && D <= ) {
string t = to_string(A) + "." + to_string(B) + "." + to_string(C) + "." + to_string(D);
if (t.size() == s.size() + ) res.push_back(t);
}
}
return res;
}
};