[leetcode] Decode Ways

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.

思路： 在解码过程中，可以选择两个字符，可以选择一个字符，所以要分别考虑，利用动态规划解决，如果选择两个字符且两个字符满足要求，且选择一个字符的时候该字符非0， 则有tmp[i] = tmp[i-1] + tmp[i-2]。但注意考虑"0*"的双字符情况。且初始边界条件也要注意，代码比较繁琐。

class Solution:    # @param s, a string    # @return an integer    def numDecodings(self, s):        if len(s) == 0:            return 0        if len(s) == 1:            if int(s) <= 0:                return 0            else:                return 1        tmp = [0 for x in range(len(s))]        if int(s[0]) != 0:            tmp[0] = 1        else:            tmp[0] = 0        if int(s[0:1]) > 26 and int(s[1]) == 0:            tmp[1] = 0        elif int(s[1]) == 0 and int(s[0:2]) <= 26:            tmp[1] = tmp[0]        elif int(s[1]) != 0 and int(s[0:2]) > 26:            tmp[1] = tmp[0]        elif int(s[1]) != 0 and int(s[0:2]) <= 26:            if s[0] == '0':                tmp[1] = 0            else:                tmp[1] = 2        for i in range(2,len(s)):            if int(s[i-1:i+1]) > 26 and int(s[i]) == 0:                tmp[i] = 0            elif int(s[i]) == 0 and int(s[i-1:i+1]) <= 26:                if s[i-1] == '0':                    tmp[i] = 0                else:                    tmp[i] = tmp[i-2]            elif int(s[i-1:i+1]) > 26 and int(s[i]) > 0:                tmp[i] = tmp[i-1]            else:                if s[i-1] == '0':                    tmp[i] = tmp[i-1]                else:                    tmp[i] = tmp[i-1] + tmp[i-2]        return tmp[len(s)-1]