Restore IP Addresses

Title:

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)

思路：使用DFS进行深度搜索。对于每个index处，考虑index~index+2，因为ip地址最多是3位整数。同时需要注意剪枝。另外根据提交，应该是不支持01.001这样的IP地址，所以需要排除

class Solution{
public:
    vector<string> restoreIpAddresses(string s) {
        vector<string> results;
        string ip;
        dfs(0,4,s,ip,results);
        return results;
    }

    void dfs(int index, int num,string s,string ip, vector<string> & results){
        if (index == s.size() && num == 0){
            results.push_back(ip);
            return ;
        }
        if (s.size() - index < num || s.size() - index > 3 * num) //剪枝
            return ;
        int sum = 0;
        for (int i = index; i < index+3; i++){
            if (i < s.size()){
                sum = sum * 10 + s[i]-'0';
                if (sum <= 255){
                    ip += s[i];
                    if (num > 1)
                        dfs(i+1,num-1,s,ip+'.',results);
                    else
                        dfs(i+1,num-1,s,ip,results);
                }
                if (sum == 0)//可以是0，但是首位是0之后，后面的就不能再用
                    break;
            }

        }
    }
};