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).

解法：注意到字符‘1’，‘2’，‘0’是和其它有些不一样的。在解码的时候会产生不同的结合方式。我们假设n个字符的解码方式又s[n]种，那么在第n+1个字符的时候，我们要考虑前一个是不是特殊字符，来确定最终的解码方式。无非就是s[n-1], s[n-1]+s[n-2],或者直接就是0.

class Solution {
public:
    int numDecodings(string s) {
        int n = s.size();
        if(n == 0) {
            return 0;
        }
        
        if(s[0] == '0') {
            return 0;
        }
        
        if(n == 1) {
            return 1;
        }
        
        vector<int> result(n, 1);
        if(s[0] == '1' && s[1] == '0' ||
          s[0] == '2' && s[1] == '0') {
            result[1] = 1;
        }
        else if(s[0] == '2' && s[1] <= '6' ||
               s[0] == '1') {
            result[1] = 2;
        }
        else if(s[0] > '2' && s[1] == '0') {
            return 0;
        }
        else {
            result[1] = 1;
        }
        
        for(int i=2; i<n; i++) {
            int lastChar = s[i-1];
            int currentChar = s[i];
            if(currentChar == '0' && (lastChar == '1' || lastChar == '2')) {
                result[i] = result[i-2];
            }
            else if(currentChar == '0') {
                return 0;
            }
            else if(lastChar == '1' ||
                   lastChar == '2' && currentChar <= '6') {
                result[i] = result[i-1] + result[i-2];
            }
            else  {
                result[i] = result[i-1];
            }
        }
        return result[n-1];
    }
};

值得注意的是，初始值为两个的时候，要把result[1]的值设置清楚，否则后面的结果肯定不对。