This problem has a lot of edge cases to be considered:

- overflow: we use a long type once it is larger than Integer.MAX_VALUE or minimum, we get over it.
- 0 sequence: because we can't have numbers with multiple digits started with zero, we have to deal with it too.
- a little trick is that we should save the value that is to be multiplied in the next recursion.

```
public class Solution {
public List<String> addOperators(String num, int target) {
List<String> rst = new ArrayList<String>();
if(num == null || num.length() == 0) return rst;
helper(rst, "", num, target, 0, 0, 0);
return rst;
}
public void helper(List<String> rst, String path, String num, int target, int pos, long eval, long multed){
if(pos == num.length()){
if(target == eval)
rst.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){
helper(rst, path + cur, num, target, i + 1, cur, cur);
}
else{
helper(rst, path + "+" + cur, num, target, i + 1, eval + cur , cur);
helper(rst, path + "-" + cur, num, target, i + 1, eval -cur, -cur);
helper(rst, path + "*" + cur, num, target, i + 1, eval - multed + multed * cur, multed * cur );
}
}
}
}
```