algorithms/decodeWays/decodeWays.cpp

// Source : https://oj.leetcode.com/problems/decode-ways/
// Author : Hao Chen
// Date   : 2014-07-16

/********************************************************************************** 
* 
* 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.
* 
*               
**********************************************************************************/

#include <iostream>
#include <string>
#include <string.h>
using namespace std;


//Count[i] = Count[i-1]              if S[i-1] is a valid char (not '0')
//         = Count[i-1]+ Count[i-2]  if S[i-1] and S[i-2] together is still a valid char (10 to 26).

int check(char ch){
    //check 0 or not
    return (!isdigit(ch) || ch=='0') ? 0 : 1;
}

int check(char ch1, char ch2){
    //check it's between 10 and 26
    return (ch1=='1' || (ch1=='2' && ch2<='6')) ? 1: 0;
}

int numDecodings(string s) {
    if (s.size()<=0) return 0;
    if (s.size()==1) return check(s[0]);
   
    int* dp = new int[s.size()];
    memset(dp, 0, s.size()*sizeof(int));

    dp[0] = check(s[0]);
    dp[1] = check(s[0]) *  check(s[1]) + check(s[0], s[1]) ;
    for (int i=2; i<s.size(); i++) {
        if (!isdigit(s[i])) break; 
        if (check(s[i])) {
            dp[i] = dp[i-1];
        }
        if (check(s[i-1], s[i])) {
            dp[i] += dp[i-2];
        }
    
    }
    int result = dp[s.size()-1];
    delete[] dp;
    return result;
}


int main(int argc, char**argv)
{
    string s = "123";
    if (argc>1){
        s = argv[1];
    }
    cout << "\"" << s << "\" : " << numDecodings(s) << endl; 
    return 0;
}