添加运算符-LintCode

给定一个仅包含数字 0 - 9 的字符串和一个目标值，返回在数字之间添加了 二元 运算符(不是一元)+, - 或 * 之后所有能得到目标值的情况。

样例：

"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 -> []

#ifndef C653_H#define C653_H#include<iostream>#include<string>#include<vector>using namespace std;class Solution {public:    /*    * @param num: a string contains only digits 0-9    * @param target: An integer    * @return: return all possibilities    */    vector<string> addOperators(string &num, int target) {        // write your code here        vector<string> res;        if (num.empty())            return res;        vector<string> dic{ "+", "-", "*" };        string str;        helper(num, str, target, 0, 0, 0, res);        return res;    }       /* str表示添加了运算符的字符串,pos表示当前位置,     * curVal,preVal分别表示当前字符串运算的总和,上一步运算得到的值     * n表示当前的数字,当运算符是加号,当前字符串运算的总和就是之前的总和加上当前数字，将本次得到的数字值置为上一步运算得到的值     * 当运算符是乘号,需要减去上一步的值再加上上一步值与n的乘积    */    void helper(string num, string str, int target, int pos, long long curVal, long long preVal, vector<string> &res)    {        if (pos == num.size()&&curVal==target)        {            res.push_back(str);            return;        }        for (int i = pos + 1; i <= num.size(); ++i)        {            string tmp = num.substr(pos,i-pos);            long long n = stoll(tmp);            if (tmp[0] == '0'&&tmp.size() > 1)                break;            if (pos == 0)                helper(num, tmp, target, i, n, n, res);            else            {                helper(num, str + "+" + tmp, target, i, curVal + n, n, res);                helper(num, str + "-" + tmp, target, i, curVal - n, -n, res);                helper(num, str + "*" + tmp, target, i, curVal - preVal + preVal*n, preVal*n, res);            }        }    }};#endif