91. Decode Ways

91. Decode Ways

1. 題目

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).

2. 思路

本題求解的是在給定的條件下(既前文中A->1，B->2)給到一個數字，好比226 有多少種解釋方法。這題能夠使用動態規劃的方法來求解。假設dp[i]爲以i爲結尾的字符串解碼方式數量的總和。那麼dp[i]的遞推規則是：

1. 若是s[i-1] != "0",則dp[i] += dp[ i - 1 ] ，表明了一次解一位，則剩下n-1個字符須要解
2. 若是一次解2位，則要求是s[i-2:i] <"27" and s[i-2:i] >'09'，dp[i] += dp[i-2]

3.實現

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]