[LeetCode] 639. Decode Ways II 解码方法之二 [LeetCode] 91. Decode Ways 解码方法 All LeetCode Questions List 题目汇总

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'.

Python:

class Solution(object):
    def numDecodings(self, s):
        """
        :type s: str
        :rtype: int
        """
        M, W = 1000000007, 3
        dp = [0] * W
        dp[0] = 1
        dp[1] = 9 if s[0] == '*' else dp[0] if s[0] != '0' else 0
        for i in xrange(1, len(s)):
            if s[i] == '*':
                dp[(i + 1) % W] = 9 * dp[i % W]
                if s[i - 1] == '1':
                    dp[(i + 1) % W] = (dp[(i + 1) % W] + 9 * dp[(i - 1) % W]) % M
                elif s[i - 1] == '2':
                    dp[(i + 1) % W] = (dp[(i + 1) % W] + 6 * dp[(i - 1) % W]) % M
                elif s[i - 1] == '*':
                    dp[(i + 1) % W] = (dp[(i + 1) % W] + 15 * dp[(i - 1) % W]) % M
            else:
                dp[(i + 1) % W] = dp[i % W] if s[i] != '0' else 0
                if s[i - 1] == '1':
                    dp[(i + 1) % W] = (dp[(i + 1) % W] + dp[(i - 1) % W]) % M
                elif s[i - 1] == '2' and s[i] <= '6':
                    dp[(i + 1) % W] = (dp[(i + 1) % W] + dp[(i - 1) % W]) % M
                elif s[i - 1] == '*':
                    dp[(i + 1) % W] = (dp[(i + 1) % W] + (2 if s[i] <= '6' else 1) * dp[(i - 1) % W]) % M
        return dp[len(s) % W]　

C++：

class Solution {
public:
    int numDecodings(string s) {
        int n = s.size(), M = 1e9 + 7;
        vector<long> dp(n + 1, 0);
        dp[0] = 1;
        if (s[0] == '0') return 0;
        dp[1] = (s[0] == '*') ? 9 : 1;
        for (int i = 2; i <= n; ++i) {
            if (s[i - 1] == '0') {
                if (s[i - 2] == '1' || s[i - 2] == '2') {
                    dp[i] += dp[i - 2];
                } else if (s[i - 2] == '*') {
                    dp[i] += 2 * dp[i - 2];
                } else {
                    return 0;
                }
            } else if (s[i - 1] >= '1' && s[i - 1] <= '9') {
                dp[i] += dp[i - 1];
                if (s[i - 2] == '1' || (s[i - 2] == '2' && s[i - 1] <= '6')) {
                    dp[i] += dp[i - 2];
                } else if (s[i - 2] == '*') {
                    dp[i] += (s[i - 1] <= '6') ? (2 * dp[i - 2]) : dp[i - 2];
                }
            } else { // s[i - 1] == '*'
                dp[i] += 9 * dp[i - 1];
                if (s[i - 2] == '1') dp[i] += 9 * dp[i - 2];
                else if (s[i - 2] == '2') dp[i] += 6 * dp[i - 2];
                else if (s[i - 2] == '*') dp[i] += 15 * dp[i - 2];
            }
            dp[i] %= M;
        }
        return dp[n];
    }
};　　

C++:

class Solution {
public:
    int numDecodings(string s) {
        long e0 = 1, e1 = 0, e2 = 0, f0, f1, f2, M = 1e9 + 7;
        for (char c : s) {
            if (c == '*') {
                f0 = 9 * e0 + 9 * e1 + 6 * e2;
                f1 = e0;
                f2 = e0;
            } else {
                f0 = (c > '0') * e0 + e1 + (c <= '6') * e2;
                f1 = (c == '1') * e0;
                f2 = (c == '2') * e0;
            }
            e0 = f0 % M;
            e1 = f1;
            e2 = f2;
        }
        return e0;
    }
};

　　

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