91. Decode Ways

A message containing letters from A-Z is being encoded to numbers using the following mapping:

'A' -> 1
'B' -> 2
...
'Z' -> 26

Given a non-empty string containing only digits, determine the total number of ways to decode it.

Example 1:

Input: "12"
Output: 2
Explanation: It could be decoded as "AB" (1 2) or "L" (12).

Example 2:

Input: "226"
Output: 3
Explanation: It could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).

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动态规划题。设s为加密字符串，dp[i]表示s[1,…,i]解密的方案数。如果$s[i]\in [1,9]$，则s[i]可以单独解密，有dp[i]+=dp[i-1]；如果$s[i-1]s[i]\in [10,26]$，则s[i]可以和s[i-1]拼起来一起解密，有dp[i]+=dp[i-2]。
DP转移公式如下：
$$\begin{cases}dp[i] += dp[i-1] & \text{if $s[i] \in [1,9]$}\\dp[i] += dp[i-2] & \text{if $s[i-1]s[i] \in [10,26]$}\\\end{cases}$$
注意如果0出现在首位则无法解密。
完整代码如下：

class Solution {
public:
    int numDecodings(string s)
    {
        if (s == "" || s[0] == ‘0’)
            return 0;
        s = "^" + s;
        int n = s.size();
        vector<int> dp(n, 0);
        dp[0] = dp[1] = 1;
        for (int i = 2; i < n; i++) {
            if (s[i] >= ‘1’ && s[i] <= ‘9’)
                dp[i] += dp[i – 1];
            if ((s[i – 1] == ‘1’ && s[i] >= ‘0’ && s[i] <= ‘9’) || (s[i – 1] == ‘2’ && s[i] >= ‘0’ && s[i] <= ‘6’))
                dp[i] += dp[i – 2];
        }
        return dp[n – 1];
    }
};

本代码提交AC，用时8MS。