[Leetcode]91. Decode Ways@python

题目

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.

题目要求

有一种信息编码方式是将字母映射成数字，’A’->1,….’Z’->26。编码后的信息有多种解码方式，本题要求给定编码后的字符串，判断有多少种解码方式。

解题思路

对此此类问题，通常用动态规划来解决。此题参考南郭子綦.
dp[i] 表示字符串前i个字符组成的子串的不同解码方式的数量
定义dp的状态方程

dp[i]=⎧⎩⎨⎪⎪dp[i−1]+dp[i−2], 10 <=s[i-2:2]<=26 and s[i−2:2]!=[10 or 20]dp[i−2],s[i-2:i] = [10 or 20]dp[i−1],others

代码

class Solution(object):
    def numDecodings(self, s):
        """
        :type s: str
        :rtype: int
        """
        if len(s) == 0 or s[0] == '0':
            return 0
        dp = [0] * (max(len(s) + 1,2))
        dp[0],dp[1] = 1,1

        for i in range(2,len(s) + 1):
            if 10 <= int(s[i - 2:i]) <= 26 and s[i - 1] != '0':
                dp[i] = dp[i - 1] + dp[i - 2]
            elif int(s[i-2:i]) == 10 or int(s[i-2:i]) == 20:
                dp[i] = dp[i - 2]
            elif s[i-1] != '0':
                dp[i] = dp[i - 1]
            else:
                return 0
        return dp[len(s)]