[LeetCode]Decode Ways II

A message containing letters from A-Z is being encoded to numbers using the following mapping way:

'A' -> 1
'B' -> 2
...
'Z' -> 26

Beyond that, now the encoded string can also contain the character '*', which can be treated as one of the numbers from 1 to 9.

Given the encoded message containing digits and the character '*', return the total number of ways to decode it.

Also, since the answer may be very large, you should return the output mod 109 + 7.

Example 1:

Input: "*"
Output: 9
Explanation: The encoded message can be decoded to the string: "A", "B", "C", "D", "E", "F", "G", "H", "I".

Example 2:

Input: "1*"
Output: 9 + 9 = 18

Note:

1. The length of the input string will fit in range [1, 105].
2. The input string will only contain the character '*' and digits '0' - '9'.

---

public class Solution {
	//*不能表示0
    static final int MOD=(int) (Math.pow(10, 9)+7);
	public int numDecodings(String s) {
		if(s.length()==0) return 0;
		long[] dp=new long[s.length()+1];
		dp[0]=1;
		if(s.charAt(0)=='0') return 0;
		else if(s.charAt(0)=='*') dp[1]=9;
		else dp[1]=1;
		for(int i=1;i<s.length();i++){
			char ch=s.charAt(i);
			char pre=s.charAt(i-1);
			if(ch=='*'){
				dp[i+1]=dp[i]*9%MOD;
				if(pre=='1'){
					dp[i+1]=(dp[i+1]+9*dp[i-1])%MOD;
				}else if(pre=='2'){
					dp[i+1]=(dp[i+1]+6*dp[i-1])%MOD;
				}else if(pre=='*'){
					dp[i+1]=(dp[i+1]+15*dp[i-1])%MOD;
				}
			}else{
				if(ch!='0') dp[i+1]=dp[i];
				if(pre=='*'){
					if(ch>='0'&&ch<='6') dp[i+1]=(dp[i+1]+2*dp[i-1])%MOD;
					else dp[i+1]=(dp[i+1]+dp[i-1])%MOD;
				}else if(pre=='1'){
					dp[i+1]=(dp[i+1]+dp[i-1])%MOD;
				}else if(pre=='2'){
					if(ch>='0'&&ch<='6') dp[i+1]=(dp[i+1]+dp[i-1])%MOD;
				}
			}
		}
		return (int)dp[dp.length-1];
    }
}