Restore IP Addresses问题及解法

问题描述：

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

示例:
Given "25525511135",

return ["255.255.11.135", "255.255.111.35"]. (Order does not matter)

问题分析：

确定是否能组成有效的ip地址，在不违反连续性的前提下，我们可以采取回溯方法，依次尝试选取一个或者两个或者三个字符组成的数来作为每个ip地址的四分之一。

过程详见代码：

class Solution {public:    vector<string> restoreIpAddresses(string s) {vector<string> res;string ip="";bs(res, ip, 0, s, 0);return res;}void bs(vector<string>& res, string& ip, int start,string s,int n){if (ip.length() == s.length() + 4){string t = ip;t.pop_back();res.push_back(t);return;}else if (n >= 4) return;int i = start;int t;            if(i < s.length())            {                t = s[i] - '0';                ip.push_back(s[i]);                ip.push_back('.');                bs(res, ip, i + 1, s,n + 1);                ip.pop_back();                ip.pop_back();            }if (i + 1 < s.length()){                 t = (s[i] - '0') * 10 + s[i + 1] - '0';                if(t >= 10)                {                    ip.push_back(s[i]);    ip.push_back(s[i + 1]);                    ip.push_back('.');                    bs(res, ip, i + 2, s, n+ 1);                    ip.pop_back();                    ip.pop_back();                    ip.pop_back();                }}if (i + 2 < s.length()){                t = (s[i] - '0') * 10 + s[i + 1] - '0';                t = t * 10 + s[i + 2] - '0';                if(t >=100 && t <=255)                {                    ip.push_back(s[i]);                    ip.push_back(s[i + 1]);                    ip.push_back(s[i + 2]);                    ip.push_back('.');                    bs(res, ip, i + 3, s, n + 1);                    ip.pop_back();                    ip.pop_back();                    ip.pop_back();                    ip.pop_back();                }    }}};