91. Decode Ways

class Solution 
{
public:
    int numDecodings(string s) 
    {
        int len=s.length();
        if(len==0||s[0]=='0')
            return 0;
        if(len==1)
            return 1;
        int res=0,count2=1,count1=1;
        for(int i=1;i<len;i++)
        {
            int cur=count1;
            if(valids(s[i])&&validd(s[i-1],s[i]))           res=count1+count2;
            else if(valids(s[i])&&!validd(s[i-1],s[i]))     res=count1;
            else if(!valids(s[i])&&validd(s[i-1],s[i]))     res=count2;
            else                                            return 0;
            count2=cur;
            count1=res;
            res=0;
        }
        return count1;
    }
    
    bool valids(char c)
    {
        return c!='0';
    }
    
    bool validd(char a,char b)
    {
        return a=='1'||(a=='2'&&b<'7');
    }
};

从头到尾扫描，和斐波那契上楼梯有点类似，问题不大