Leetcode Restore IP Addresses

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)

方法一：暴力破解法

写代码时没有思虑周全，一开始忘了对每个字段进行判断小于255，之后调试发现，需要处理当数字位数大于1但以0为开头的情况。

class Solution {public:    vector<string> restoreIpAddresses(string s) {        vector<string> result;        int len = s.length();        int a,b,c,d;        for(a=1;a<=3;a++)        for(b=1;b<=3;b++)        for(c=1;c<=3;c++)        for(d=1;d<=3;d++)        {            if(a+b+c+d == len)            {                string strA = s.substr(0,a);                string strB = s.substr(a,b);                string strC = s.substr(a+b,c);                string strD = s.substr(a+b+c,d);                int IntA = stoi(s.substr(0,a));                int IntB = stoi(s.substr(a,b));                int IntC = stoi(s.substr(a+b,c));                int IntD = stoi(s.substr(a+b+c,d));                if(IntA <= 255 && IntB <= 255 && IntC <= 255 && IntD <= 255)                {                    if((a != 1 && strA[0] == '0') || (b != 1 && strB[0] == '0') || (c != 1 && strC[0] == '0')                    || (d != 1 && strD[0] == '0'))                        continue;                    else                    {                         string temp = strA +'.'+ strB+'.' + strC+'.' + strD;                         result.push_back(temp);                    }                }            }        }        return result;    }};

方法二：递归方法

restore函数中的result需要加引用，不然只会是拷贝，找这个bug蛋疼了蛮久 - -

class Solution {public:    vector<string> restoreIpAddresses(string s) {        vector<string> result;        restore(s,result,0,0,"");        return result;    }        void restore(string s,vector<string>& result,int idx,int count,string restored)    {        if(count > 4) return;        if(count == 4 && idx == s.length())            result.push_back(restored);            for(int i=1;i<4;i++)            {                if(idx+i > s.length()) break;                string temp = s.substr(idx,i);                if(stoi(temp) > 255) continue;                if(i > 1 && temp[0] == '0') continue;                restore(s,result,idx+i,count+1,restored+temp+(count==3?"":"."));            }    }};