91. Decode Ways

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).

狀態轉移方程：

dp[i] = dp[i - 1]             if s[i - 1] != '0'
      + dp[i - 2]             if 10 <= stoi(s.substr(i - 2, 2)) <= 26

class Solution {
public:
    int numDecodings(string s) {
        int n = s.size();
        // dp 數列代表的是 s 的前 i 個位置可以被解碼的次數
        vector<int> dp(n + 1);
        
        // 起始值為 1，第 0 位可以被解碼
        dp[0] = 1;

        // 0 開頭的數字不能被解碼，所以要算 0 次
        // 其他數字開頭的話，算 1 次
        dp[1] = (s[0] == '0' ? 0 : 1);
        
        // 迭代從第 2 個位置開始
        // 也就是 s = "12" 的話，會從下標的地方開始算
        //             ^
        for(int i = 2; i < n + 1; i++) {
            // 考慮一個數字的情況，如果前一個位置不等於 0
            // 延續上一個可以解碼的次數：dp[i] = dp[i - 1]
            if(s[i - 1] != '0')
                dp[i] = dp[i - 1];

            // 考慮兩個數字的情況
            int twoDigits = stoi(s.substr(i - 2, 2));

            // 如果符合 10 ~ 26 之間，加上 dp[i - 2] 的結果
            if(10 <= twoDigits && twoDigits <= 26) {
                dp[i] += dp[i - 2];
            }
        }
        return dp[n];
    }
};

• 時間複雜度：
  $O(N)$
• 空間複雜度：
  $O(N)$