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71 lines
2.3 KiB
Python
71 lines
2.3 KiB
Python
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#coding: utf-8
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''' mbinary
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#######################################################################
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# File : dfs.py
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# Author: mbinary
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# Mail: zhuheqin1@gmail.com
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# Blog: https://mbinary.xyz
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# Github: https://github.com/mbinary
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# Created Time: 2019-05-27 10:02
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# Description:
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from leetcode-cn #1048: https://leetcode-cn.com/problems/longest-string-chain/
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给出一个单词列表，其中每个单词都由小写英文字母组成。
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如果我们可以在 word1 的任何地方添加一个字母使其变成 word2，那么我们认为 word1 是 word2 的前身。例如，"abc" 是 "abac" 的前身。
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词链是单词 [word_1, word_2, ..., word_k] 组成的序列，k >= 1，其中 word_1 是 word_2 的前身，word_2 是 word_3 的前身，依此类推。
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从给定单词列表 words 中选择单词组成词链，返回词链的最长可能长度。
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#######################################################################
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'''
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class Solution:
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def longestStrChain(self, words: List[str]) -> int:
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def isAdj(s1,s2):
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if len(s1)>len(s2):
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s1,s2 = s2,s1
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n1,n2 = len(s1),len(s2)
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if n2-n1!=1:
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return False
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i=j=0
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flag = False
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while i<n1 and j<n2:
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if s1[i]!=s2[j]:
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if flag:
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return False
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flag = True
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j+=1
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else:
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i+=1
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j+=1
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return True
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def dfs(begin):
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ans = 1
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w = words[begin]
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n = len(w)
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if n+1 in lenDic:
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for nd in lenDic[n+1]:
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#print(w,words[nd],isAdj(w,words[nd]))
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if isAdj(w,words[nd]):
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ans = max(ans,1+dfs(nd))
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return ans
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lenDic = {}
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for i in range(len(words)):
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n = len(words[i])
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if n in lenDic:
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lenDic[n].add(i)
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else:
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lenDic[n]={i}
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lens = sorted(lenDic)
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n = len(lens)
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ans = 0
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for i in range(n):
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if ans < n-i:
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for nd in lenDic[lens[i]]:
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ans = max(ans,dfs(nd))
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return ans
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