My easily understanding solution and explanation to it!


  • 0

    Yesterday, Bro Luo told me: "Why don't you improve your English by writing your blogs in English?" I think it may be a good way and i did so today.
    Problem Description:

       A message containing letters from A - Z  is being encoded to numbers using the follow 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.

    My Solution:

     I used dynamic programming to solve it. For each char in string s can be divided in four types. 
    

    1.Only can be combined with pre-one

    thus dp[i] = dp[i - 2], cause the pre one(i - 1) must combine with i

    2.Only can be self alone

    thus dp[i] = dp[i - 1]

    1. Both can

    thus dp[i] = dp[i - 1] + dp[i - 2]

    4 Both can NOT:

    thus return 0, cause the message can be decode

     Of course, return dp[len - 1] if not return before.
    class Solution {
    

    public:
    bool canCombine(char a, char b)
    {
    if(a >= '3') return false;
    if(a == '2'){
    if(b <= '6') return true;
    else return false;
    }
    if(a == '0') return false;
    return true;
    }

    bool canAlone(char a)
    {
        if(a >= '1' && a <= '9') return true;
        return false;
    }
    
    int numDecodings(string s) {
        if(s == "") return 0;
        if(s[0] == '0') return 0;
        if(s.length() < 2) return 1;
        int dp[50000];
        dp[0] = 1;
        bool canComb = canCombine(s[0],s[1]);
        bool canAlon = canAlone(s[1]);
        if(canComb && canAlon)
            dp[1] = dp[0] + 1;
        else if(!canComb && !canAlon)
            return 0;
        else
            dp[1] = dp[0];
        for(int i = 2; i < s.length(); i ++){
            bool canComb = canCombine(s[i - 1], s[i]);
            bool canAlon = canAlone(s[i]);
            if(canComb && canAlon){
                dp[i] = dp[i - 1] + dp[i - 2];
            }else if(canComb){
                dp[i] = dp[i - 2];
            }else if(canAlon){
                dp[i] = dp[i - 1];
            }else{
                dp[i] = 0;
                return 0;
            }
        }
        return dp[s.length() - 1];
    }
    

    };


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