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.

解决方法：利用动态规划来做，设ｄｐ［ｉ］为前ｉ个数的译码方法，ｓ［ｌｅｎ］为字符串。

首先当ｌｅｎ＝０或者ｓ［０］＝０时，是不存在译码方法的；

ｄｐ［０］＝０，

ｄｐ［ｉ］＝ｓ［ｉ－１］单独作为一个字母的译码方法＋ｓ［ｉ－１］和ｓ［ｉ－２］一起作为一个字母的译码方法，１＜＝ｉ＜＝ｌｅｎ．

具体代码为：

class Solution {public:    int numDecodings(string s) {        int len=s.size();        int i,j,result;        int *dp=new int[len+1];        if(len==0||s[0]=='0')        return 0;        dp[0]=1;                for(i=1;i<=len;i++){            if(s[i-1]=='0')              dp[i]=0;            else               dp[i]=dp[i-1];            if(i>1&&(s[i-2]=='1'||(s[i-2]=='2'&&s[i-1]<='6')))              dp[i]+=dp[i-2];        }        result=dp[len];        delete []dp;        return result;    }};