[LeetCode]Decode Ways

class Solution {
//DFS here will be TLE
//So the intuitive is that this can be solved by DP
//get the maximum or minimum or number, we should give DP a shot for these kind of problems
//transform equation:
//if(IsCode(i)) f[i]+=f[i-1]
//if(IsCode(i-1, i)) f[i]+=f[i-2]
//initialize: f[0]=1, f[1]=IsCode(s[0])
public:
	bool IsCode(string s, int start, int end)
	{
		if(end-start+1 >= 3) return false;
		if(s[start] == '0') return false;
		int res = 0;
		for (int i = start; i <= end; ++i)
			res = res*10+s[i]-'0';
		if(res >= 1 && res <= 26)
			return true;
	}
	int numDecodings(string s) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		int n = s.size();
		if(n == 0) return 0;//note here
		vector<int> f(n+1, 0);
		f[0] = 1;
		if( IsCode(s, 0, 0) ) f[1] = 1;//when use constant subscript, always remember will it out of range?
		for (int i = 2; i <= n; ++i)
		{
			if( IsCode(s, i-1, i-1) ) f[i] += f[i-1];
			if( IsCode(s, i-1-1, i-1) ) f[i] += f[i-2];
		}
		return f[n];
	}
};

second time

class Solution {
public:
    int numDecodings(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(s.size() == 0) return 0;
        vector<int> f(s.size()+1, 0);
        f[0] = 1;
        if(s[0] != '0') f[1] = 1;
        for(int i = 2; i <= s.size(); ++i)
        {
            int curDigit = s[i-1]-'0';
            int prevDigit = s[i-2]-'0';
            int sum = prevDigit*10+curDigit;
            if(sum >= 10 && sum <= 26) f[i] += f[i-2];
            if(curDigit != 0) f[i] += f[i-1];
        }
        return f[s.size()];
    }
};