Restore IP Addresses

题目：

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)
算法思想：
这道题考察的是回溯算法（backtracing），一个合法的ip地址包含4个小于等于255的整数，回溯算法的过程是：先找到第一个，再找第二个，再找第三个，当第四个不满足条件时，重新找第三个...当所有情况都被找完时，结束。

回溯算法本质上是一个树的前序遍历，所以有两个写法：

1.递归算法：因为是树，所以递归最简单，也最容易理解。

2.迭代算法：这个稍微麻烦些，考虑的边界条件也较多，但效率最高。

void helper(string segment, int level, string temp, vector &result)
{
if (segment.size() == 0)
return;

if (level < 4)
{
for (int i = 0; i < 3; i++)
{
if (i+1 >= segment.size() || stoi(segment.substr(0, i+1)) > 255)
return;

helper(segment.substr(i+1, segment.size()-i-1), level+1, temp+segment.substr(0, i+1)+'.', result);

if (segment[0] == '0')
return;
}
}
else
{
if (segment[0] == '0' && segment.size() > 1)
return;
if (segment.size() > 3 || stoi(segment) > 255)
return;

temp += segment;
result.push_back(temp);
}
}

vector restoreIpAddresses(string s)
{
vector res;

helper(s, 1, "", res);

return res;
}

vector restoreIpAddresses(string s)
{
vector res;
stringstream ss;
list plist;

if (s.length() < 4)
return res;

plist.push_back(0);
while (true)
{
while (plist.size() < 5)
plist.push_back(plist.back()+1);

ss.str("");
if(plist.back() == s.length())
{
list::iterator listiter = plist.begin();
listiter++;
for (int i = 0; i < s.length(); i++)
{
ss << s[i];
if (i != s.length()-1 && i+1 == *listiter)
{
ss << '.';
listiter++;
}
}

res.push_back(ss.str());
}

int k;
while (plist.size() != 1)
{
k = plist.back();
plist.pop_back();

if (k-plist.back() < 3 && k+5-plist.size() <= s.length() && s[plist.back()] != '0'
&& stoi(s.substr(plist.back(), k-plist.back()+1)) <= 255)
{
plist.push_back(k+1);
break;
}
}

if (plist.size() == 1)
return res;
}
}