DecodeWays.h

/*
 Author:     Annie Kim, anniekim.pku@gmail.com
 Date:       May 20, 2013
 Problem:    Decode Ways
 Difficulty: Easy
 Source:     http://leetcode.com/onlinejudge#question_91
 Notes:
 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.

 Solution: 1. dp. Note that '0' must be decoded in together with the number in front of it.
           2. dp. Time : O(n); Space : O(1).
 */

class Solution {
public:
    int numDecodings_1(string s) {
        if (s.empty() || s[0] == '0') 
            return 0;
        int N = s.size();
        int dp[N+1];
        dp[0] = 1;
        dp[1] = 1;
        for (int i = 1; i < N; ++i)
        {
            if (s[i] == '0')
            {
                if (s[i-1] != '1' && s[i-1] != '2')
                    return 0;
                dp[i+1] = dp[i-1];
            }
            else
            {
                int num = s[i] - '0' + (s[i-1] - '0') * 10;
                if (num >= 11 && num <= 26)
                    dp[i+1] = dp[i] + dp[i-1];
                else
                    dp[i+1] = dp[i];
            }
        }
        return dp[N];
    }
    int numDecodings_2(string s) {
        if (s.empty() || s[0] == '0') return 0;
        int N = s.size();
        int f0 = 1, f1 = 1;
        for (int i = 1; i < N; ++i) {
            if (s[i] == '0') f1 = 0;
            int num = s[i] - '0' + (s[i-1] - '0') * 10;
            if (num < 10 || num > 26) {
                f0 = 0;
            }
            int tmp = f1;
            f1 = f1 + f0;
            f0 = tmp;
        }
        return f1;
    }
};