给表达式添加运算符

给定一个仅包含数字 0-9 的字符串 num 和一个目标值整数 target ，在 num 的数字之间添加 二元 运算符（不是一元）+、- 或 * ，返回所有能够得到目标值的表达式。

示例 1:

输入: num = "123", target = 6
输出: ["1+2+3", "1*2*3"] 

示例 2:

输入: num = "232", target = 8
输出: ["2*3+2", "2+3*2"]

示例 3:

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

示例 4:

输入: num = "00", target = 0
输出: ["0+0", "0-0", "0*0"]

示例 5:

输入: num = "3456237490", target = 9191
输出: []

class Solution {
    int n;
    String num;
    List<String> ans;
    int target;
    public List<String> addOperators(String num, int target) {
        this.n = num.length();
        this.num = num;
        this.target = target;
        this.ans = new ArrayList<String>();
        StringBuffer expr = new StringBuffer();
        dfs(expr, 0, 0, 0);
        return ans;
    }
    public void dfs(StringBuffer sba, long sum, long prepareMultiply, int index) {
        StringBuffer sb = new StringBuffer(sba);
        if (index == n) {
            if (sum == target) {
                ans.add(sb.toString());
            }
            return;
        }
        int sign = sb.length();
        if (index > 0) {
            sb.append("0");
        }
        long val = 0;
        for (int i = index; i < n && (i == index || num.charAt(index) != '0'); i++) {
            val = val * 10 + (num.charAt(i) - '0');
            sb.append(num.charAt(i));
            if (index == 0) {
                dfs(sb, val, val, i + 1);
                continue;
            }
            sb.setCharAt(sign, '+');
            dfs(sb, sum + val, val, i + 1);
            sb.setCharAt(sign, '-');
            dfs(sb, sum - val, -val, i + 1);
            sb.setCharAt(sign, '*');
            dfs(sb, sum - prepareMultiply + prepareMultiply * val, prepareMultiply * val, i + 1);
        }
    }
}