【重点】【DP】剑指offer46_new：解密数字

题目
思路：

Python

class Solution:
    def crackNumber(self, ciphertext: int) -> int:
        s = str(ciphertext)
        n = len(s)
        dp = [0] * (n+1) # dp[i]前i位数字有几种解密结果
        dp[0], dp[1] = 1, 1
        for i in range(2, n+1):
            if int(s[i-2: i]) >= 10 and int(s[i-2: i]) <= 25:
                dp[i] = dp[i-2] + dp[i-1]
            else:
                dp[i] = dp[i-1]
        return dp[n]

Java

class Solution {
    public int crackNumber(int ciphertext) {
        String numStr = String.valueOf(ciphertext);
        int n = numStr.length();
        int dp[] = new int[n + 1]; // dp[n]表示前n个数字的解密方法数量
        dp[0] = 1;
        dp[1] = 1;
        for (int i = 2; i <= n; ++i) {
            int num1 = numStr.charAt(i - 2) - '0';
            int num2 = numStr.charAt(i - 1) - '0';
            int tmpVal = 10 * num1 + num2;
            if (tmpVal >= 10 && tmpVal <= 25) {
                dp[i] = dp[i - 1] + dp[i -2];
            } else {
                dp[i] = dp[i - 1];
            }
        }

        return dp[n];
    }
}