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 an encoded message containing digits, determine the total number of ways to decode it.

For example,
Given encoded message “12”, it could be decoded as “AB” (1 2) or “L” (12).

The number of ways decoding “12” is 2.

动态规划：
dp[i] 表示i的字符的合法编码数量，取决于当前字符
dp[i] += dp[i-1],if s[i-1] ~(‘0’–‘9’)
dp[i] += dp[i-2],if s[i-2] == ‘1’ || (s[i-2]==’2’ && s[i-1]>=’0’ && s[i-1]<=’6’)

    int numDecodings(string s) {
        if(s.empty() || s[0] == '0')
            return 0;
        int n = s.size();
        vector<int> dp(n+1, 0);
        dp[0] = 1;
        dp[1] = 1;
        for(int i = 2; i <= n; i++){
            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]>='0' && s[i-1]<='6'))
                dp[i] += dp[i-2]; 
        }
        return dp[n];
    }