[leetcode] 91. Decode Ways @ python

原题

A message containing letters from A-Z is being encoded to numbers using the following mapping:

‘A’ -> 1
‘B’ -> 2
…
‘Z’ -> 26
Given a non-empty string containing only digits, determine the total number of ways to decode it.

Example 1:

Input: “12”
Output: 2
Explanation: It could be decoded as “AB” (1 2) or “L” (12).
Example 2:

Input: “226”
Output: 3
Explanation: It could be decoded as “BZ” (2 26), “VF” (22 6), or “BBF” (2 2 6).

解法

动态规划. 关键部分是推导动态转移方程, 用dp[i] 表示到i点时的解码方法的个数, 那么

dp[i] = dp[i-1] if s[i] != "0"
       +dp[i-2] if "10" <= s[i-1:i+1] <= "26"

此时有edge case: 当s长度<= 2时, 例如s = ‘12’, 会出现错误, 因此为了方便推导, 将dp头部增加一个长度, dp[0] = 1, dp[i]为s[:i]的解码方法个数. 此时动态转移方程变为

dp[i]  = dp[i-1] if s[i-1] != "0"
        +dp[i-2] if i!= 1 and "10" <= s[i-2:i] <= "26"

Time: O(n)
Space: O(n)

代码

class Solution(object):
    def numDecodings(self, s):
        """
        :type s: str
        :rtype: int
        """
        # base case
        if not s: return 0
        dp = [0]*(len(s) + 1)
        dp[0] = 1
        for i in range(1, len(dp)):
            if s[i-1] != '0':
                dp[i] += dp[i-1]
            if i!= 1 and '10' <= s[i-2:i] <= '26':
                dp[i] += dp[i-2]
                
        return dp[-1]