[leetcode 282]Expression Add Operators

Given a string that contains only digits 0-9 and a target value, return all possibilities to add binary operators (not unary) +, -, or * between the digits so they evaluate to the target value.

Examples: 

"123", 6 -> ["1+2+3", "1*2*3"] 
"232", 8 -> ["2*3+2", "2+3*2"]
"105", 5 -> ["1*0+5","10-5"]
"00", 0 -> ["0+0", "0-0", "0*0"]
"3456237490", 9191 -> []

Credits:
Special thanks to  for adding this problem and creating all test cases.

给定一个由数字字符组成的字符串，在任意字符之间加入+、-、*构成数学表达式，使得表达式的结果为所给的target

AC代码：

class Solution
{
public:
    vector<string> addOperators(string num, int target)
    {
        vector<string> res;
        string temp="";
        int len=num.size();
        for(int i=1; i<=len; ++i)
        {
            string curString=num.substr(0,i);
            if(curString.size()>1&&curString[0]=='0')
                continue;
            string nextString=num.substr(i,len-i);
            long long curNum=stringToLonglong(curString);
            temp="";
            cal(nextString,target,curNum,curNum,curString,res);
        }
        return res;
    }

private:

    long long stringToLonglong(string s)
    {
        stringstream ss;
        ss<<s;
        long long  res;
        ss>>res;
        return res;
    }


    void cal(string num,int target,long long last,long long cur,string temp,vector<string> &res)
    {
        int len=num.size();
        if(len==0&&target==cur)
        {
            res.push_back(temp);
            return ;
        }
        for(int i=1; i<=len; ++i)
        {
            string curString=num.substr(0,i);
            string nextString=num.substr(i,len-i);
            if(curString.size()>1&&curString[0]=='0')
                return ;
            long long curNum=stringToLonglong(curString);
            cal(nextString,target,curNum,cur+curNum,temp+"+"+curString,res);
            cal(nextString,target,-curNum,cur-curNum,temp+"-"+curString,res);
            cal(nextString,target,last*curNum,cur-last+last*curNum,temp+"*"+curString,res);
        }
    }
};

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