递归+必要剪枝

class Solution{
public:
	bool isValid(string s){
		if(s.size()>3||!s.size()||s.size()>1&&s[0]==‘0‘)return false;
		int sum=0;
		for(int i=0;i<s.size();i++){
			sum*=10;
			sum+=s[i]-‘0‘;
		}
		return sum>=0&&sum<=255;
	}
	void dfs(string &s,int start,int count,string path,vector<string> &result){
	    if(count>3)return;
		if(count==3&&isValid(s.substr(start))){
			path.append(s.substr(start));
			result.push_back(path);
			return;
		}
		for(int i=start;i<s.size()&&i-start<3;i++){
			string sub=s.substr(start,i-start+1);
			if(isValid(sub)){
				string t=path;
				path.append(sub+‘.‘);
				dfs(s,i+1,count+1,path,result);
				path=t;
			}
		}
	}
	vector<string> restoreIpAddresses(string s){
		vector<string> result;
		dfs(s,0,0,"",result);
		return result;
	}
};

LeetCode OJ:Restore IP Addresses