639. 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'.

Approach #1: DP. [C++]

class Solution {
    int mod = 1000000007;
    
    public int numDecodings(String s) {
        if (s.isEmpty()) return 0;
        long[] dp = new long[2];
        dp[0] = 1;
        dp[1] = ways(s.charAt(0));
        // int ans = 0;
        for (int i = 1; i < s.length(); ++i) {
            long ans = ways(s.charAt(i)) * dp[1] + ways(s.charAt(i-1), s.charAt(i)) * dp[0];
            ans %= mod;
            dp[0] = dp[1];
            dp[1] = ans;
        }
        return (int)dp[1];
    }
    
    public int ways(char c) {
        if (c == '*') return 9;
        if (c == '0') return 0;
        return 1;
    }
    
    public int ways(char c1, char c2) {
        if (c1 == '*' && c2 == '*') return 15;
        if (c1 == '*') 
            if (c2 >= '0' && c2 <= '6') return 2;
            else return 1;
        else if (c2 == '*') 
            if (c1 == '1') return 9;
            else if (c1 == '2') return 6;
            else return 0;
        else {
            int num = (c1 - '0') * 10 + (c2 - '0');
            if (num >= 10 && num <= 26) return 1;
            else return 0;
        }

    }
}

　　

Reference:

http://zxi.mytechroad.com/blog/dynamic-programming/leetcode-639-decode-ways-ii/