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  • 91. Decode Ways

    91. Decode Ways

    1. 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 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

    本题求解的是在给定的条件下(既前文中A->1,B->2)给到一个数字,比如226 有多少种解释方法。这题可以使用动态规划的方法来求解。假设dp[i]为以i为结尾的字符串解码方式数量的总和。那么dp[i]的递推规则是:

    1. 如果s[i-1] != "0",则dp[i] += dp[ i - 1 ] ,代表了一次解一位,则剩下n-1个字符需要解
    2. 如果一次解2位,则要求是s[i-2:i] <"27" and s[i-2:i] >'09',dp[i] += dp[i-2]

    3.实现

    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]
    
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  • 原文地址:https://www.cnblogs.com/bush2582/p/10926870.html
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