题目大意：
给你一个加密的数字字符串，问你一共有多少种不同的解密方式。
Problem:
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.
Idea:
Dynamic Programming
Solution:
C++
Time complexity: O(n^2)
Space complexity: O(n^2)
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// Author: Huahua // Runtime: 6 ms class Solution { public: int numDecodings(string s) { if(s.length() == 0) return 0; m_ways[""] = 1; return ways(s); } private: int ways(const string& s) { if (m_ways.count(s)) return m_ways[s]; if (s[0] == '0') return 0; if (s.length() == 1) return 1; int w = ways(s.substr(1)); const int prefix = stoi(s.substr(0, 2)); if (prefix <= 26) w += ways(s.substr(2)); m_ways[s] = w; return w; } unordered_map<string, int> m_ways; }; |
C++
Time complexity: O(n)
Space complexity: O(n)
|
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// Author: Huahua // Runtime: 3 ms class Solution { public: int numDecodings(string s) { if(s.length() == 0) return 0; return ways(s, 0, s.length() - 1); } private: int ways(const string& s, int l, int r) { if (m_ways.count(l)) return m_ways[l]; if (s[l] == '0') return 0; if (l >= r) return 1; // Single digit or empty. int w = ways(s, l + 1, r); const int prefix = (s[l] - '0') * 10 + (s[l + 1] - '0'); if (prefix <= 26) w += ways(s, l + 2, r); m_ways[l] = w; return w; } // Use l as key. unordered_map<int, int> m_ways; }; |
C++
Time complexity: O(n)
Space complexity: O(1)
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class Solution { public: int numDecodings(string s) { if (s.empty() || s[0] == '0') return 0; if (s.length() == 1) return 1; const int n = s.length(); int w1 = 1; int w2 = 1; for (int i = 1; i < n; ++i) { int w = 0; if (!isValid(s[i]) && !isValid(s[i - 1], s[i])) return 0; if (isValid(s[i])) w = w1; if (isValid(s[i - 1], s[i])) w += w2; w2 = w1; w1 = w; } return w1; } private: bool isValid(const char c) { return c != '0'; } bool isValid(const char c1, const char c2) { const int num = (c1 - '0')*10 + (c2 - '0'); return num >= 10 && num <= 26; } }; |
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