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# [10/1] 91. Decode Ways

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### 본문

Medium

A message containing letters from `A-Z` can be encoded into numbers using the following mapping:

```'A' -> "1"
'B' -> "2"
...
'Z' -> "26"
```

To decode an encoded message, all the digits must be grouped then mapped back into letters using the reverse of the mapping above (there may be multiple ways). For example, `"11106"` can be mapped into:

• `"AAJF"` with the grouping `(1 1 10 6)`
• `"KJF"` with the grouping `(11 10 6)`

Note that the grouping `(1 11 06)` is invalid because `"06"` cannot be mapped into `'F'` since `"6"` is different from `"06"`.

Given a string `s` containing only digits, return the number of ways to decode it.

The test cases are generated so that the answer fits in a 32-bit integer.

Example 1:

```Input: s = "12"
Output: 2
Explanation: "12" could be decoded as "AB" (1 2) or "L" (12).
```

Example 2:

```Input: s = "226"
Output: 3
Explanation: "226" could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).
```

Example 3:

```Input: s = "06"
Output: 0
Explanation: "06" cannot be mapped to "F" because of the leading zero ("6" is different from "06").
```

Constraints:

• `1 <= s.length <= 100`
• `s` contains only digits and may contain leading zero(s).
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Submissions
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## 학부유학생님의 댓글

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``````class Solution:
def numDecodings(self, s: str) -> int:
if int(s[0]) == 0: return 0
dp = [0]*(len(s)+1)
dp[0:2] = [1,1]

for i in range(2, len(s) + 1):
if int(s[i-1]) > 0 :
dp[i] += dp[i-1]
if 10 <= int(s[i-2:i]) <= 26:
dp[i] += dp[i-2]

return dp[-1]``````
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