Leetcode 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'.

 

 

 
 
 1 public class Solution {
 2     public int NumDecodings(string s) {
 3         if (s.Length == 0 || s[0] == '0') return 0;
 4         
 5         long dp0 = 1, dp1 = s[0] == '*' ? 9 : 1;
 6         
 7         for (int i = 1; i < s.Length; i++)
 8         {
 9             long dp2 = 0;
10             
11             if (s[i - 1] == '*')
12             {
13                 if (s[i] == '*')
14                 {
15                     dp2 = dp0 * (9 + 6) + 9 * dp1;
16                 }
17                 else if ((int)s[i] <= (int)'6')
18                 {
19                     dp2 = dp0 * 2 + (s[i] == '0' ? 0 : dp1);
20                 }
21                 else
22                 {
23                     dp2 = dp0 + dp1;
24                 }
25             }
26             else
27             {
28                 if (s[i] == '*')
29                 {
30                     if (s[i - 1] == '1')
31                     {
32                         dp2 = dp0 * 9 + dp1 * 9;
33                     }
34                     else if (s[i - 1] == '2')
35                     {
36                         dp2 = dp0 * 6 + dp1 * 9;
37                     }
38                     else
39                     {
40                         dp2 = dp1 * 9;
41                     }
42                 }
43                 else
44                 {            
45                     var decode = ((int)s[i - 1] - (int)'0') * 10 + ((int)s[i] - (int)'0');
46                     if (decode >= 10 && decode <= 26)
47                     {
48                         dp2 = dp0;
49                     } 
50 
51                     if (decode % 10 != 0)
52                     {
53                         dp2 += dp1;
54                     }       
55                 }
56 
57             }
58             
59             dp0 = dp1 % (1000000007);
60             dp1 = dp2 % (1000000007);
61         }
62         
63         return (int)(dp1 % (1000000007));
64     }
65 }

 

posted @ 2017-11-13 12:46  逸朵  阅读(417)  评论(0)  编辑  收藏  举报