[leetcode]282. Expression Add Operators

题目链接：https://leetcode.com/problems/expression-add-operators/description/

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 -> []

class Solution {private:    // cur: {string} expression generated so far.    // pos: {int}    current visiting position of num.    // cv:  {long}   cumulative value so far.    // pv:  {long}   previous operand value.    // op:  {char}   previous operator used.    void dfs(std::vector<string>& res, const string& num, const int target, string cur, int pos, const long cv, const long pv, const char op) {        if (pos == num.size() && cv == target) {            res.push_back(cur);        } else {            for (int i=pos+1; i<=num.size(); i++) {                string t = num.substr(pos, i-pos);                long now = stol(t);                if (to_string(now).size() != t.size()) continue;                dfs(res, num, target, cur+'+'+t, i, cv+now, now, '+');                dfs(res, num, target, cur+'-'+t, i, cv-now, now, '-');                dfs(res, num, target, cur+'*'+t, i, (op == '-') ? cv+pv - pv*now : ((op == '+') ? cv-pv + pv*now : pv*now), pv*now, op);            }        }    }public:    vector<string> addOperators(string num, int target) {        vector<string> res;        if (num.empty()) return res;        for (int i=1; i<=num.size(); i++) {            string s = num.substr(0, i);            long cur = stol(s);            if (to_string(cur).size() != s.size()) continue;            dfs(res, num, target, s, i, cur, cur, '#');         // no operator defined.        }        return res;    }};