91. Decode Ways

动态规划问题
创建一个长度为n+1的数组来储存子问题的结果.
状态转移方程:
dp[i] =
dp[i-1]当s[i] != "0"
+dp[i-2] 当"09" < s[i-1:i+1] < "27"
数组最后一个数即为返回的结果.
基础情况:
dp[0] = 1表示一个空字符串只有一种方式去decode
当字符串的第一个字符不为0时,dp[1]为1,否则为0.

class Solution(object):
    def numDecodings(self, s):
        """
        :type s: str
        :rtype: int
        """
        if s == '':
            return 0
        dp = [0 for x in range(len(s)+1)]
        dp[0] = 1
        for i in range(1, len(s)+1):
            if s[i-1] != "0":
                dp[i] += dp[i-1]
            if i != 1 and s[i-2:i] < "27" and s[i-2:i] > "09":
                dp[i] += dp[i-2]
        return dp[-1]