639. Decode Ways II

class Solution {
public:
    int numDecodings(string s) {
        int n = s.length();
        if (n == 0) return 0;
        vector<long> dp(n+1, 0);
        dp[0] = 1;
        for (int i = 0; i < n; i++) {
            if (s[i] == '*') {
                dp[i+1] += dp[i] * 9;
                if (i > 0) {
                    if (s[i-1] == '*' || s[i-1] == '1')
                        dp[i+1] += dp[i-1] * 9;
                    if (s[i-1] == '*' || s[i-1] == '2')
                        dp[i+1] += dp[i-1] * 6;
                }
            }
            else {
                if (s[i] != '0')
                    dp[i+1] += dp[i];
                if (i > 0) {
                    if (s[i-1] == '*' || s[i-1] == '1')
                        dp[i+1] += dp[i-1];
                    if ((s[i-1] == '*' || s[i-1] == '2') && s[i] <= '6')
                        dp[i+1] += dp[i-1];
                }
            }
            dp[i+1] %= 1000000007;
        }
        return dp[n];
    }
};

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转载自www.cnblogs.com/JTechRoad/p/9121224.html