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71 lines
2.3 KiB
Python
71 lines
2.3 KiB
Python
#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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