[LeetCode] Decode Ways

最新推荐文章于 2025-12-21 18:32:56 发布

转载最新推荐文章于 2025-12-21 18:32:56 发布 · 61 阅读

文章标签：

#c/c++

本文深入探讨了如何使用动态规划解决字符串解码问题，具体阐述了解码算法的核心思想、实现步骤及实例分析。通过具体示例，展示了如何计算给定编码消息的所有可能解码方式，为理解字符串编码与解码提供了实用的解决方案。

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.

DP解决

 1 class Solution {
 2 private:
 3     vector<int> f;
 4 public:
 5     int getF(int index)
 6     {
 7         if (index < 0)
 8             return 1;
 9         else
10             return f[index];
11     }
12     
13     int numDecodings(string s) {
14         // Start typing your C/C++ solution below
15         // DO NOT write int main() function
16         if (s.size() == 0)
17             return 0;
18             
19         f.resize(s.size());
20         
21         for(int i = 0; i < s.size(); i++)
22         {
23             f[i] = 0;
24             if (i >= 1)
25             {
26                 string a(s, i - 1, 2);
27                 if ("10" <= a && a <= "26")
28                     f[i] += getF(i-2);
29                 if ('1' <= s[i] && s[i] <= '9')
30                     f[i] += getF(i-1);
31             }
32             else
33             {
34                 if ('1' <= s[i] && s[i] <= '9')
35                     f[i] = 1;
36             }
37         }
38         
39         return f[f.size() - 1];
40     }
41 };