19.2.23 [LeetCode 91] Decode Ways

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).

题意

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

题解

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 };

原文地址：https://www.cnblogs.com/yalphait/p/10423833.html