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.
Example
Given encoded message 12, it could be decoded as AB (1 2) or L (12).
The number of ways decoding 12 is 2.
LeetCode上的原题，请参见我之前的博客Decode Ways。
解法一：
class Solution { public: /** * @param s a string, encoded message * @return an integer, the number of ways decoding */ int numDecodings(string& s) { if (s.empty()) return 0; int n = s.size(); vector<int> dp(n + 2, 1); for (int i = 2; i < n + 2; ++i) { if (s[i - 2] == '0') dp[i] = 0; else dp[i] = dp[i - 1]; if (i >= 3 && (s[i - 3] == '1' || (s[i - 3] == '2' && s[i - 2] <= '6'))) { dp[i] += dp[i -2]; } } return dp.back(); } };
解法二：
class Solution { public: /** * @param s a string, encoded message * @return an integer, the number of ways decoding */ int numDecodings(string& s) { if (s.empty()) return 0; vector<int> dp(s.size() + 1, 0); dp[0] = 1; for (int i = 1; i < dp.size(); ++i) { if (s[i - 1] >= '1' && s[i - 1] <= '9') dp[i] += dp[i - 1]; if (i >= 2 && s.substr(i - 2, 2) <= "26" && s.substr(i - 2, 2) >= "10") { dp[i] += dp[i - 2]; } } return dp.back(); } };
解法三：
class Solution { public: /** * @param s a string, encoded message * @return an integer, the number of ways decoding */ int numDecodings(string& s) { if (s.empty() || s.front() == '0') return 0; int c1 = 1, c2 = 1; for (int i = 1; i < s.size(); ++i) { if (s[i] == '0') c1 = 0; if (s[i - 1] == '1' || (s[i - 1] == '2' && s[i] <= '6')) { c1 = c1 + c2; c2 = c1 - c2; } else { c2 = c1; } } return c1; } };
本文转自博客园Grandyang的博客，原文链接：解码方法[LintCode] Decode Ways ，如需转载请自行联系原博主。
