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.

解题思路：倒序dp。假设要求字符串ABCXXXXX。如果已知BCXXXXX的decode方式数量，加入一个A之后，如果AB是[11,26]之内，那么ABCXXXXX的编码方式=（A，BCXXXXX的编码方式）+（AB，CXXXXX）的编码方式。否则ABCXXXXX的编码方式=（A，BCXXXXX的编码方式）

class Solution {
public:
    int numDecodings(string s) {
        int n=s.length();
        if(n==0)return 0;
        int dp[n+1]={0};
        dp[n]=1;
        for(int i=n-1;i>=0;i--){
            if(s[i]==‘0‘){
                dp[i]=0;continue;
            }
            if(i!=n-1&&s[i]==‘1‘||s[i]==‘2‘&&s[i+1]<=‘6‘)
                dp[i]=dp[i+1]+dp[i+2];
            else dp[i]=dp[i+1];
        }
        return dp[0];
    }
};

91. Decode Ways