Decode ways [leetcode] DP

Question: How many possible results are there to convert the number string s into a letter?

DP (I): number of possible results for S [0... I-1. We set Si = I-1

Analyze s [Si]:

1. s [Si-1] = '1' | s [Si-1] = '2' & S [Si] = '6'

S [0... si] can be converted into letters corresponding to s [Si] + s [0... si-1] corresponding letter, may also be s [Si-1 .. si] corresponding letter + s [0... si-2] corresponding letters

That is, dp (I) = dp (I-1) + dp (I-2)

2. s [Si] = '0', which is easy to ignore.

2.1 s [Si-1] = '1' | s [Si-1] = '2'

S [0... Si] can be converted to s [Si-1... Si] corresponding letter + s [0... Si-2] corresponding letter

DP (I) = dp (I-2)

2.2 s [Si-1] does not exist or is another number

DP (I) = 0;

3. Other cases

DP (I) = dp (I-1)

The Code is as follows:

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

Decode ways [leetcode] DP