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]
3. 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.
My code:
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]; } };
原文：http://www.cnblogs.com/luntai/p/5621887.html