力扣-回溯-93 复原IP地址

思路

用一个vector存放可能的结果，然后用一个变量判断插入点的数量，假设再最后一段后也插入点

代码

class Solution {
public:
    vector<string> result;
    vector<string> path;

    int toNum(string s){
        int d = 1;
        int result = 0;
        for(int i = s.size()-1; i >= 0; i--){
            result += ( (char)s[i] - '0' ) * d;
            d *= 10;
        }
        return result;
    }

    void backTracking(string s, int startIndex, int dotNum){
        if(startIndex == s.size() && dotNum == 4){
            string ip;
            for(int i = 0; i < path.size(); i++){
                ip += path[i];
                if(i != path.size() - 1) ip += ".";
            }
            result.push_back(ip);
            return;
        }

        string num_string; 
        for(int i = startIndex; i < s.size(); i++){
            num_string += s[i];
            int num = toNum(num_string);
            if(to_string(num) != num_string) break; 
            if(num > 255) break;
            path.push_back(to_string(num));
            backTracking(s, i+1, dotNum+1);
            path.pop_back();
        }
        return;
    }
    vector<string> restoreIpAddresses(string s) {
        backTracking(s, 0, 0);

        return result;
    }
};