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:
- The length of the input string will fit in range [1, 105].
- 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/
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