算法精粹(algorithm-essentials)

感谢soulmachine@github提供内容
## Decode Ways


### 描述

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.


### 分析

跟第???节[Climbing Stairs](../linear-list/array/climbing-stairs.md)很类似,不过多加一些判断逻辑。


### 代码

{% if book.java %}
```java
// Decode Ways
// 动规,时间复杂度O(n),空间复杂度O(1)
public class Solution {
public int numDecodings(String s) {
if (s.isEmpty() || s.charAt(0) == '0') return 0;

int prev = 0;
int cur = 1;
// 长度为n的字符串,有 n+1个阶梯
for (int i = 1; i <= s.length(); ++i) {
if (s.charAt(i-1) == '0') cur = 0;

if (i < 2 || !(s.charAt(i - 2) == '1' ||
(s.charAt(i - 2) == '2' && s.charAt(i - 1) <= '6')))
prev = 0;

int tmp = cur;
cur = prev + cur;
prev = tmp;
}
return cur;
}
}
```
{% endif %}

{% if book.cpp %}
```cpp
// Decode Ways
// 动规,时间复杂度O(n),空间复杂度O(1)
class Solution {
public:
int numDecodings(const string &s) {
if (s.empty() || s[0] == '0') return 0;

int prev = 0;
int cur = 1;
// 长度为n的字符串,有 n+1个阶梯
for (size_t i = 1; i <= s.size(); ++i) {
if (s[i-1] == '0') cur = 0;

if (i < 2 || !(s[i - 2] == '1' ||
(s[i - 2] == '2' && s[i - 1] <= '6')))
prev = 0;

int tmp = cur;
cur = prev + cur;
prev = tmp;
}
return cur;
}
};
```
{% endif %}


### 相关题目

* [Climbing Stairs](climbing-stairs.md)