I used a dp array of size n + 1 to save subproblem solutions. `dp[0]`

means an empty string will have one way to decode, `dp[1]`

means the way to decode a string of size 1. I then check one digit and two digit combination and save the results along the way. In the end, `dp[n]`

will be the end result.

```
public class Solution {
public int numDecodings(String s) {
if(s == null || s.length() == 0) {
return 0;
}
int n = s.length();
int[] dp = new int[n+1];
dp[0] = 1;
dp[1] = s.charAt(0) != '0' ? 1 : 0;
for(int i = 2; i <= n; i++) {
int first = Integer.valueOf(s.substring(i-1, i));
int second = Integer.valueOf(s.substring(i-2, i));
if(first >= 1 && first <= 9) {
dp[i] += dp[i-1];
}
if(second >= 10 && second <= 26) {
dp[i] += dp[i-2];
}
}
return dp[n];
}
}
```