# 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: []
提示：
1 <= num.length <= 10num 仅含数字-231 <= target <= 231 - 1
## template
```java
class Solution {
int n;
String num;
List ans;
int target;
public List addOperators(String num, int target) {
this.n = num.length();
this.num = num;
this.target = target;
this.ans = new ArrayList();
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);
}
}
}
```
## 答案
```java
```
## 选项
### A
```java
```
### B
```java
```
### C
```java
```