# Strobogrammatic Number II

247. Strobogrammatic Number II

A strobogrammatic number is a number that looks the same when rotated 180 degrees (looked at upside down).

Find all strobogrammatic numbers that are of length = n.

For example,

Given n = 2, return ["11","69","88","96"].

n = 0: none

n = 1: 0, 1, 8

n = 2: 11, 69, 88, 96

n = 3: 101, 609, 808, 906, 111, 619, 818, 916, 181, 689, 888, 986

n = 4: 1001, 6009, 8008, 9006, 1111, 6119, 8118, 9116, 1691, 6699, 8698, 9696, 1881, 6889, 8888, 9886, 1961, 6969, 8968, 9966

```
public class Solution {
public List<String> findStrobogrammatic(int n) {
return helper(n, n);
}
List<String> helper(int n, int m) {
if (n == 0) {
return new ArrayList<String>(Arrays.asList(""));
}
if (n == 1) {
return new ArrayList<String>(Arrays.asList("0", "1", "8"));
}
List<String> list = helper(n - 2, m);
List<String> result = new ArrayList<String>();
for (String str : list) {
if (n != m) {
result.add("0" + str + "0");
}
result.add("1" + str + "1");
result.add("6" + str + "9");
result.add("8" + str + "8");
result.add("9" + str + "6");
}
return result;
}
}
```

```
public class Solution {
public List<String> findStrobogrammatic(int n) {
List<String> one = Arrays.asList("0", "1", "8");
List<String> two = Arrays.asList("");
List<String> result = two;
if (n % 2 == 1) {
result = one;
}
for (int i = (n % 2) + 2; i <= n; i += 2) {
List<String> newList = new ArrayList<>();
for (String str : result) {
if (i != n) {
newList.add("0" + str + "0");
}
newList.add("1" + str + "1");
newList.add("6" + str + "9");
newList.add("8" + str + "8");
newList.add("9" + str + "6");
}
result = newList;
}
return result;
}
}
```

Hope this helps,

Michael