LeetCode OJ 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.

I have met a same problem in Sicily..But this version is a little difficult cause there may be some case which is illegal especially for zero number..

class Solution {public:    inline bool is_letter(char front, char back) {        if (front >= '3' || front == '0') return false;        else if (front == '1') return true;        else if (back > '6') return false;        else return true;    }    int numDecodings(string s) {        if (s[0] == '0') return 0;  // leading zero is illegal        for (int i = 1; i < s.size(); i++) if (!is_letter(s[i - 1], s[i]) && s[i] == '0') return 0;  // if there is a zero like 30, if divide them, the single zero is illegal, if not, the 30 is illegal        int ans[s.size() + 1];        for(int i = 0; i < s.size() + 1; ans[i++] = 0);        ans[0] = 1;        if (is_letter(s[0], s[1]) && s[1] != '0') ans[1] = 2;  // notice that zero cannot be a letter alone        else ans[1] = 1;        for (int i = 2; i < s.size(); i++)            if (s[i] == '0') ans[i] = ans[i - 2];  // then zero must combine  with s[i - 1] to form a letter, the s[i - 1] is fixed(it must be combine with zero or this will be illegal)            else if (is_letter(s[i - 1], s[i])) ans[i] = ans[i - 2] + ans[i - 1];  // if s[i - 1] and s[i] can combine to stand for a letter, ans[i] = ans[i - 1](not combine) + ans[i - 2](combine)            else ans[i] = ans[i - 1];        return ans[s.size() - 1];    }};