[LeetCode]46. Restore IP Addresses复原IP地址

Given a string containing only digits, restore it by returning all possible valid IP address combinations.

For example:
Given

"25525511135"

,

return

["255.255.11.135", "255.255.111.35"]

. (Order does not matter)

Subscribe to see which companies asked this question

解法1：暴力破解法。三层循环确定前面三个数字，最后剩下的所有作为第四个数字。实际每层循环最多执行三遍。循环中使用作为正确IP地址的条件进行剪枝。需要注意考虑各种边界条件。

class Solution {
public:
vector<string> restoreIpAddresses(string s) {
int n = s.size();
string addr = "";
vector<string> res;
for (int i = 0; i < n - 3 && i < 3; ++i)
{
string fir(s.begin(), s.begin() + i + 1);
if ((fir[0] != '0' || fir.size() == 1) && stoi(fir) <= 255)
{
for (int j = i + 1; j < n - 2 && j < i + 4; ++j)
{
string sec(s.begin() + i + 1, s.begin() + j + 1);
if ((sec[0] != '0' || sec.size() == 1) && stoi(sec) <= 255)
{
for (int k = j + 1; k < n - 1 && k < j + 4; ++k)
{
string trd(s.begin() + j + 1, s.begin() + k + 1);
if ((trd[0] != '0' || trd.size() == 1) && stoi(trd) <= 255)
{
string fth(s.begin() + k + 1, s.end());
if ((fth[0] != '0' || fth.size() == 1) && fth.size() < 4 && stoi(fth) <= 255)
{
addr += fir + '.' + sec + '.' + trd + '.' + fth;
res.push_back(addr);
addr.clear();
}
}
}
}
}
}
}
return res;
}
};