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.

"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 -> []

public class Solution {
    public List<String> addOperators(String num, int target) {
        List<String> res = new ArrayList<String>();
        dfs(num, target, res, "", 0, 0, 0);
        return res;
    }
    
    public void dfs(String num, int target, List<String> res, String path, int pos, long eval, long mult){
        if(pos == num.length()){
            if(eval == target)
                res.add(path);
            return;
        }
        
        for(int i = pos; i < num.length(); i++){
            if(i != pos && num.charAt(pos) == '0') break;
            long cur = Long.parseLong(num.substring(pos, i+1));
            if(pos == 0){
                dfs(num, target, res, path + cur, i+1, cur, cur);
            } else {
                dfs(num, target, res, path + "+" + cur, i+1, eval + cur, cur);
                dfs(num, target, res, path + "-" + cur, i+1, eval - cur, -cur);
                /*
                 唯一需要注意的就是乘法的情况，"345" 产生出 3+4*5， 
                 到达5的时候，eval = 7, mul 4
                 eval-mult 算出不包含4的值，这里是7-4=3， 4是乘号以前的算式值，4*5, 算乘法部分。
                 参考Basic calculator II
                */
                dfs(num, target, res, path + "*" + cur, i+1, (eval-mult) + mult*cur , mult*cur);
            }
        }
    }
}