leetcode1048

 1 class Solution:
 2     def __init__(self):
 3         self.dp = {}
 4 
 5     def backtrack(self,word,subword):
 6         self.dp[word] = max(self.dp[word],self.dp[subword]+1)
 7         for j in range(len(subword)):
 8             subword2 = subword[:j] + subword[j+1:]
 9             if subword2 in self.dp.keys() and self.dp[subword2] == 0:
10                 self.backtrack(subword,subword2)
11 
12     def longestStrChain(self, words: 'List[str]') -> int:
13         n = len(words)
14         words = sorted(words,key=lambda x:len(x))
15         for i in range(n):
16             word = words[i]
17             self.dp[word] = 0
18 
19         for word in self.dp.keys():
20             for j in range(len(word)):
21                 subword = word[:j] + word[j+1:]
22                 if subword in self.dp.keys():
23                     if self.dp[subword] == 0:
24                         self.backtrack(word,subword)
25                     else:
26                         self.dp[word] = max(self.dp[word],self.dp[subword]+1)
27             if self.dp[word] == 0:
28                 self.dp[word] = 1
29         return max(self.dp.values())

本题是思路是使用动态规划，dp是一个字典，其含义是：假设以某个string作为链的结尾时形成的链的最大长度。

因此最后dp的values中，最大的值就是输入列表所能形成的最长的链的长度。

 

来源：https://www.cnblogs.com/asenyang/p/10928401.html