[LeetCode26]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.

Analysis:

The key to this problem is : num(current) = num(current-1) + num(current-2).
Be careful with the conditions of above equation:
if the current char =='0'
if the current char and previous char >26
if the previous char =='0'

This can be think as a simple DP problem, res[i]= res[i-1] if s[i] is valid + res[i-2] if s[i-1:i] is valid.

Java

public int numDecodings(String s) {
        if(s.length()<=0) return 0;
        int len = s.length();
        int res[] = new int[len];
        if(s.charAt(0)!='0') res[0] = 1;
        if(len==1) return res[0];
        if(check(s.substring(0, 2))) res[1]++;
        if(s.charAt(0)!='0'&&s.charAt(1)!='0') res[1]++;
        //DP
        for(int i=2;i<len;i++){
        	if(s.charAt(i)!='0')res[i] +=res[i-1];
        	if(check(s.substring(i-1, i+1)))
        		res[i]+=res[i-2];
        }
        return res[len-1];
    }
	public boolean check(String string){
		if(string.length()==0) return false;
		if(string.charAt(0)=='0') return false;
		if(string.length()==2){
			if(string.charAt(0)>'2'||(string.charAt(0)=='2'&&string.charAt(1)>'6'))
				return false;
		}
		return true;
	}

c++

int check(char one){
    return (one != '0') ? 1:0;
}
int check(char one, char two){
    return (one =='1' || (one=='2'&& two<='6'))?1:0;
}
int numDecodings(string s) {
    if(s.empty() || s[0]=='0') return 0;
    if(s.size()==1) return check(s[0]);
    int fn=0, fn_1 = 0,fn_2 = 1;
    fn_1 = (check(s[0])*check(s[1]))+check(s[0],s[1]);
    for(int i=2;i<s.size();i++){
        if(check(s[i])) fn+=fn_1;
        if(check(s[i-1],s[i])) fn+=fn_2;
        if(fn == 0)
            return 0;
        fn_2 = fn_1;
        fn_1 = fn;
        fn = 0;
    }
    return fn_1;
}