Decode Ways II

Description

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 10^9 + 7.

1. The length of the input string will fit in range [1, 10^5].
2. The input string will only contain the character * and digits 0 - 9.

   public class Solution {
       /**
        * @param s: a message being encoded
        * @return: an integer
        */
        public int numDecodings(String s) {
           if (s == null || s.length() == 0) {
               return 0;
           }
   
           final int mod = 1000000007;
           int n = s.length();
           int[] f = new int[n + 1];
           f[0] = 1;
           for (int i = 1; i <= n; i++) {
               f[i] = 0;
               if (s.charAt(i - 1) == ‘*‘) {
                   f[i] = (int)((f[i] + 9L * f[i - 1]) % mod);
                   if (i >= 2) {
                       if (s.charAt(i - 2) == ‘*‘) {
                           f[i] = (int)((f[i] + 15L * f[i - 2]) % mod);
                       }
                       else if (s.charAt(i - 2) == ‘1‘) {
                           f[i] = (int)((f[i] + 9L * f[i - 2]) % mod);
                       }
                       else if (s.charAt(i - 2) == ‘2‘) {
                           f[i] = (int)((f[i] + 6L * f[i - 2]) % mod);
                       }
                   }
               }
               else {
                   if (s.charAt(i - 1) != ‘0‘) {
                       f[i] = (f[i] + f[i - 1]) % mod;
                   }
                   if (i >= 2) {
                       if (s.charAt(i - 2) == ‘*‘){
                           if (s.charAt(i - 1) <= ‘6‘) {
                               f[i] = (int)((f[i] + 2L * f[i - 2]) % mod);
                           }
                           else {
                               f[i] = (f[i] + f[i - 2]) % mod;
                           }
                       }
                       else {
                           int twoDigits = (s.charAt(i - 2) - ‘0‘) * 10 + s.charAt(i - 1) - ‘0‘;
                           if (twoDigits >= 10 && twoDigits <= 26) {
                               f[i] = (f[i] + f[i - 2]) % mod;
                           }
                       }
                   }
               }
           }
   
           return f[n];
       }
   }
   

   　　

原文地址：https://www.cnblogs.com/FLAGyuri/p/12078271.html