[LeetCode] Restore IP Addresses 回溯

標籤：

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)

 

Hide Tags Backtracking String 　　   這題是一道回溯題，其實記錄好斷開的位置便容易處理了。我使用的便是遞迴搜尋的方法，用一個數組記錄了斷開的位置。

#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;}

 

   

[LeetCode] Restore IP Addresses 回溯