Problem:

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’.

Idea:

DP

Time complexity: O(n)

Space complexity: O(1)

Solution:

C++

// Author: Huahua
// Runtime: 58 ms
class Solution {
public:
    int numDecodings(string s) {
        if (s.empty()) return 0;        
        //           dp[-1]  dp[0]
        long dp[2] = {1, ways(s[0])};
        for (int i = 1; i < s.length(); ++i) {
            long dp_i = ways(s[i]) * dp[1] + ways(s[i - 1], s[i]) * dp[0];
            dp_i %= kMod;
            dp[0] = dp[1];
            dp[1] = dp_i;
        }
        return dp[1];
    }
private:
    static constexpr int kMod = 1000000007;    
    
    int ways(char c) {
        if (c == '0') return 0;
        if (c == '*') return 9;
        return 1;
    }
    
    int ways(char c1, char c2) {
        if (c1 == '*' && c2 == '*') 
            return 15;
        if (c1 == '*') {
          return (c2 >= '0' && c2 <= '6') ? 2 : 1;
        } else if (c2 == '*') {
            switch (c1) {
                case '1': return 9;
                case '2': return 6;
                default: return 0;
            }
        } else {
            int prefix = (c1 - '0') * 10 + (c2 - '0');
            return prefix >= 10 && prefix <= 26;
        }        
    }
};

Related Problems:

• [解题报告] LeetCode 91. Decode Ways

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