标签：ESS git bst min ping put win 输出 determine
A message containing letters from A-Z is being encoded to numbers using the following mapping:
‘A‘ -> 1
‘B‘ -> 2
...
‘Z‘ -> 26
Given a non-empty string containing only digits, determine the total number of ways to decode it.
Example 1:
Input: "12"
Output: 2
Explanation: It could be decoded as "AB" (1 2) or "L" (12).
Example 2:
Input: "226"
Output: 3
Explanation: It could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).
输出一串数字可能的解读方式，这串数字的解读只能由1~26组成
记忆性递归，好像挺慢的
1 class Solution { 2 public: 3 vector<int>mark; 4 int cnt(string s) { 5 int one = s[0] - ‘0‘, two = 0, ans = 0, l = s.length(); 6 if (one == 0)return 0; 7 if (mark[l] != -1)return mark[l]; 8 if (l > 1)two = one * 10 + s[1] - ‘0‘; 9 else if (l == 0)return 1; 10 ans += cnt(s.substr(1)); 11 if (two <= 26 && two >= 1) 12 ans += cnt(s.substr(2)); 13 mark[l] = ans; 14 return ans; 15 } 16 int numDecodings(string s) { 17 int l = s.length(); 18 mark = vector<int>(l + 1, -1); 19 return cnt(s); 20 } 21 };
于是dp，o(n)的
1 class Solution { 2 public: 3 int numDecodings(string s) { 4 int l = s.length(); 5 vector<int>dp(l + 1, 0); 6 dp[0] = 1; 7 if (s[0] - ‘0‘ != 0)dp[1] = 1; 8 for (int i = 1; i < l; i++) { 9 int one = s[i - 1] - ‘0‘, two = (s[i] - ‘0‘) + one * 10; 10 if (one != 0 && two <= 26 && two >= 1) 11 dp[i+1] +=dp[i - 1]; 12 if(s[i]!=‘0‘) 13 dp[i+1] += dp[i]; 14 } 15 return dp[l]; 16 } 17 };
19.2.23 [LeetCode 91] Decode Ways
标签：ESS git bst min ping put win 输出 determine
原文地址：https://www.cnblogs.com/yalphait/p/10423833.html
