[LintCode] Restore IP Address 复原IP地址

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, 4, "", res);
return res;
}
void helper(string s, int k, string out, vector<string>& res) {
if (k == 0) {
if (s.empty()) res.push_back(out);
return;
}
for (int i = 1; i < 4; ++i) {
if (s.size() < i) break;
int val = stoi(s.substr(0, i));
if (val > 255 || i != to_string(val).size()) continue;
helper(s.substr(i), k - 1, out + s.substr(0, i) + (k == 1 ? "" : "."), 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 = 1; a < 4; ++a)
for (int b = 1; b < 4; ++b)
for (int c = 1; c < 4; ++c)
for (int d = 1; d < 4; ++d)
if (a + b + c + d == s.size()) {
int A = stoi(s.substr(0, 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 <= 255 && B <= 255 && C <= 255 && D <= 255) {
string t = to_string(A) + "." + to_string(B) + "." + to_string(C) + "." + to_string(D);
if (t.size() == s.size() + 3) res.push_back(t);
}
}
return res;
}
};