leetcode1048

class Solution:
    def __init__(self):
        self.dp = {}

    def backtrack(self,word,subword):
        self.dp[word] = max(self.dp[word],self.dp[subword]+1)
        for j in range(len(subword)):
            subword2 = subword[:j] + subword[j+1:]
            if subword2 in self.dp.keys() and self.dp[subword2] == 0:
                self.backtrack(subword,subword2)

    def longestStrChain(self, words: 'List[str]') -> int:
        n = len(words)
        words = sorted(words,key=lambda x:len(x))
        for i in range(n):
            word = words[i]
            self.dp[word] = 0

        for word in self.dp.keys():
            for j in range(len(word)):
                subword = word[:j] + word[j+1:]
                if subword in self.dp.keys():
                    if self.dp[subword] == 0:
                        self.backtrack(word,subword)
                    else:
                        self.dp[word] = max(self.dp[word],self.dp[subword]+1)
            if self.dp[word] == 0:
                self.dp[word] = 1
        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。