Leetcode [91]: Decode ways

最新推荐文章于 2021-05-26 22:15:12 发布

OString24

最新推荐文章于 2021-05-26 22:15:12 发布

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分类专栏： programming Leetcode 文章标签： dfs string

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本文深入探讨了将字母从A到Z编码为数字1至26的算法，并通过实例展示了如何计算给定数字字符串的不同解码方式总数。文章提供了三种解码算法实现：递归深度优先搜索、使用内存缓存中间结果以及动态规划方法。

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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).

91.Decode Ways

dfs way:

class Solution(object):
    def numDecodings(self, s):
        """
        :type s: str
        :rtype: int
        """
        out=[]
        self.ret=0
        
        def helper(b):
            if b==len(s):
                self.ret+=1
                return 

            for i in range(b,len(s)):
                sub=s[b:i+1]
                if i-b+1>2 or int(sub)<1 or int(sub)>26:
                    break
                out.append(sub)
                helper(i+1)
                out.pop()
        helper(0)   
        return self.ret

using memory to cache middle result

out=[]
        seen={}
        def helper(b):
            if b in seen:
                return seen[b]
            
            if b==len(s):
                return 1
            
            ret=0
            for i in range(b,len(s)):
                if i-b+1>2:
                    break
                sub=s[b:i+1]
                if int(sub)<1 or int(sub)>26:
                    break
                #out.append(sub)
                ret+=helper(i+1)
                #out.pop()
            seen[b]=ret
            return ret
        
        return helper(0)

using dp

#
# @lc app=leetcode id=91 lang=python
#
# [91] Decode Ways
#
class Solution(object):
    def numDecodings(self, s):
        """
        :type s: str
        :rtype: int
        """
        if not s or s[0]=='0':
            return 0

        dp=[0 for i in range(len(s)+1)]
        dp[0]=1 #
        for i in range(1,len(s)+1):
            if s[i-1]=='0':
                dp[i]=0
            else:
                dp[i]=dp[i-1]

            if i>1 and (s[i-2]=='1' or (s[i-2]=='2' and s[i-1]<='6')):
                dp[i]+=dp[i-2]

        return dp[len(s)]