[LeetCode] Restore IP Addresses 回溯

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)

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   这题是一道回溯题，其实记录好断开的位置便容易处理了。我使用的便是递归搜索的方法，用一个数组记录了断开的位置。

#include <string>
#include <vector>
#include <iostream>
using namespace std;


class Solution {
public:
    vector<string> restoreIpAddresses(string s) {
        vector<string> ret;
        if(s.size()<4)  return ret;
        int idx[4] = {0,0,0,0};
        helpFun(ret,s,idx,1);
        return ret;
    }

    void helpFun(vector<string> &ret,string & s,int * idx, int id)
    {
        if(id==4){
//            for(int i =0;i<4;i++)
//                cout<<idx[i]<<" ";
//            cout<<endl;
            if(helpFun2(s.substr(idx[3])))
                ret.push_back(s.substr(idx[0],idx[1]-idx[0])+"."+
                              s.substr(idx[1],idx[2]-idx[1])+"."+
                              s.substr(idx[2],idx[3]-idx[2])+"."+
                              s.substr(idx[3]) );
            return ;
        }
        for(int i =idx[id-1]+1;i<s.length();i++){
            if(helpFun2(s.substr(idx[id-1],i-idx[id-1]))){
                idx[id] = i;
                helpFun(ret,s,idx,id+1);
            }
            else
                return ;
        }
    }

    bool helpFun2(string s)
    {
        if(s.length()==1&&s[0]=='0')    return true;
        if(s[0]=='0')   return false;
        int sum = 0;
        for(int i=0;i<s.length();i++){
            sum = sum* 10 + s[i]-'0';
            if(sum>255) return false;
        }
        return true;
    }
};

int main()
{
    string s="25525511135";
    Solution sol;
    vector<string> ret = sol.restoreIpAddresses(s);
    for(int i=0;i<ret.size();i++)
        cout<<ret[i]<<endl;
    return 0;
}