[Leetcode Series] [] [medium] decoding algorithm method

topic:

Topic links:  https://leetcode-cn.com/problems/decode-ways/

 

Problem-solving ideas:

Dynamic Programming

If the last digit is not 0, only 1 decoding mode

Starting from the penultimate position, assuming index value index, a total of the following three cases:

1. s [index] = 0, skip the current bit continue traversing
2. 0 <s [index] <= 26, the current position is valid, then the bit index decoder decodes index + cumulative manner to Embodiment 1
3. 0 <s [index: index + 2] <= 26, the current bit + a valid bit, the bit index additive decoding method of decoding scheme index + 2

 

Code:

class Solution:
    def numDecodings(self, s: str) -> int:
        # 此处dp数组初始化长度 + 1,原因如下：
        # 如果最后一位是0,直接跳过,不需要叠加多余出来的一项
        # 如果最后一位不是0,是有效值,相当于对最后一位初始化为了0
        # 如果倒数第二位 + 倒数第一位是有效值,则不用添加特殊处理，与其他情况处理方式相同
        # dp[index] = dp[index + 1] + dp[index + 2]
        # 避免了倒数第二位index + 2时下标溢出
        dp = [0] * (len(s) + 1)
        dp[-1] = 1
        for index in range(len(s) - 1, -1, -1):
            double_str = 0
            if index < len(s) - 1:
                double_str = int(s[index:index + 2])

            if '0' == s[index]:
                continue
            elif 0 < double_str <= 26:
                dp[index] += dp[index + 1] + dp[index + 2]
            else:
                dp[index] += dp[index + 1]
                
        return dp[0]