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Simple DP.
Scanning from the end. If current is 0, then no ways as "0", but it can be treated as "10", "20". So do another check whether it can has other combinations from i+2.
But if the last one is "0", there is no way to decode it except len-2 is "1" or "2". So assign one more space to dp hold 1 way for special case.
1 class Solution { 2 public: 3 int numDecodings(string s) { 4 int len = s.size(); 5 if (len == 0) return 0; 6 vector<int> dp(len+1, 1); 7 for (int i = len-1; i >= 0; i--) { 8 if (s[i] == ‘0‘) dp[i] = 0; 9 else dp[i] = dp[i+1]; 10 11 if (i < len-1 && (s[i] == ‘1‘ || (s[i] == ‘2‘ && s[i+1] <= ‘6‘))) dp[i] += dp[i+2]; 12 } 13 return dp[0]; 14 } 15 };
LeetCode – Refresh – Decode Ways
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原文地址：http://www.cnblogs.com/shuashuashua/p/4349365.html
