[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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Backtracking String

 　　

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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()<)  return ret;
        int idx[] = {,,,};
        helpFun(ret,s,idx,);
        return ret;
    }
 
    void helpFun(vector<string> &ret,string & s,int * idx, int id)
    {
        if(id==){
//            for(int i =0;i<4;i++)
//                cout<<idx[i]<<" ";
//            cout<<endl;
            if(helpFun2(s.substr(idx[])))
                ret.push_back(s.substr(idx[],idx[]-idx[])+"."+
                              s.substr(idx[],idx[]-idx[])+"."+
                              s.substr(idx[],idx[]-idx[])+"."+
                              s.substr(idx[]) );
            return ;
        }
        for(int i =idx[id-]+;i<s.length();i++){
            if(helpFun2(s.substr(idx[id-],i-idx[id-]))){
                idx[id] = i;
                helpFun(ret,s,idx,id+);
            }
            else
                return ;
        }
    }
 
    bool helpFun2(string s)
    {
        if(s.length()==&&s[]=='')    return true;
        if(s[]=='')   return false;
        int sum = ;
        for(int i=;i<s.length();i++){
            sum = sum*  + s[i]-'';
            if(sum>) return false;
        }
        return true;
    }
};
 
int main()
{
    string s="";
    Solution sol;
    vector<string> ret = sol.restoreIpAddresses(s);
    for(int i=;i<ret.size();i++)
        cout<<ret[i]<<endl;
    return ;
}