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.

Example 1:

Input: num = "123", target = 6
Output: ["1+2+3", "1*2*3"] 

Example 2:

Input: num = "232", target = 8
Output: ["2*3+2", "2+3*2"]

Example 3:

Input: num = "105", target = 5
Output: ["1*0+5","10-5"]

Example 4:

Input: num = "00", target = 0
Output: ["0+0", "0-0", "0*0"]

Example 5:

Input: num = "3456237490", target = 9191
Output: []

void backtracking(const string& num, int target, string path, vector<string>& res, int pos, long val, long pre_multi) {
    if (pos == num.length()) {
        if (val == target)
            res.emplace_back(path);
        return;
    }
    for (int i = pos; i < num.length(); ++i) {
        // Skip a number starting with '0' digit
        if (i != pos && num[pos] == '0') break;
        string cur_num = num.substr(pos, i + 1 - pos);
        long cur = stol(cur_num);
        if (pos == 0)
            backtracking(num, target, cur_num, res, i + 1, cur, cur);
        else {
            backtracking(num, target, path + "+" + cur_num, res, i + 1, val + cur, cur);
            backtracking(num, target, path + "-" + cur_num, res, i + 1, val - cur, -cur);
            backtracking(num, target, path + "*" + cur_num, res, i + 1, val - pre_multi + pre_multi * cur, pre_multi * cur);
        }
    }
}
vector<string> addOperators(string num, int target) { // time: O(4^n); space: O(n)
    if (num.empty()) return {};
    vector<string> res;
    backtracking(num, target, "", res, 0, 0, 0);
    return res;
}