leetcode 1048. Longest String Chain（python）

描述

You are given an array of words where each word consists of lowercase English letters. wordA is a predecessor of wordB if and only if we can insert exactly one letter anywhere in wordA without changing the order of the other characters to make it equal to wordB.

• For example, "abc" is a predecessor of "abac", while "cba" is not a predecessor of "bcad".

A word chain is a sequence of words [word1, word2, ..., wordk] with k >= 1, where word1 is a predecessor of word2, word2 is a predecessor of word3, and so on. A single word is trivially a word chain with k == 1. Return the length of the longest possible word chain with words chosen from the given list of words.

Example 1:

Input: words = ["a","b","ba","bca","bda","bdca"]
Output: 4
Explanation: One of the longest word chains is ["a","ba","bda","bdca"].

Example 2:

Input: words = ["xbc","pcxbcf","xb","cxbc","pcxbc"]
Output: 5
Explanation: All the words can be put in a word chain ["xb", "xbc", "cxbc", "pcxbc", "pcxbcf"].

Example 3:

Input: words = ["abcd","dbqca"]
Output: 1
Explanation: The trivial word chain ["abcd"] is one of the longest word chains.
["abcd","dbqca"] is not a valid word chain because the ordering of the letters is changed.

Note:

1 <= words.length <= 1000
1 <= words[i].length <= 16
words[i] only consists of lowercase English letters.

解析

根据题意，给定一个单词数组，其中每个单词由小写英文字母组成。

wordA 是 wordB 的前身当且仅当我们可以在 wordA 的任意位置准确插入一个字母而不改变其他字符的顺序以使其等于 wordB。

• 例如，“abc”是“abac”的前身，而“cba”不是“bcad”的前身。

一个词链是一个词序列 [word1, word2, ..., wordk] ，其中 k >= 1，其中 word1 是 word2 的前身，word2 是 word3 的前身，以此类推。返回从给定单词列表中选择的单词的最长可能单词链的长度。

这是一个典型的动态规划题目，因为这条链有长度限制，所以我们先将 words 按照顺序进行升序排序，这样我们就能在长度限制的情况下，使用两重循环挨个找子数组能构成的最长的单词链长度。我们定义一个一维数组 dp ，dp[i] 表示的是以索引 i 单词为末尾的单词链最长的长度，如果两个单词 words[j] 和 words[i] 符合题意，那么就更新 dp[i]= max(dp[i], dp[j]+1) 。遍历结束直接返回 dp 中的最大值即可。

时间复杂度为 O(16*N^2) ，空间复杂度为 O(N^2) 。

解答

class Solution(object):
    def longestStrChain(self, words):
        """
        :type words: List[str]
        :rtype: int
        """
        def check(s1, s2):
            M, N = len(s1), len(s2)
            if M + 1 != N:
                return False
            i, j = 0, 0
            while i < M and j < N:
                if s1[i] == s2[j]: i += 1
                j += 1
            return i == M

        words.sort(key=lambda x: (len(x), x))
        N = len(words)
        dp = [1] * N  
        for i in range(N):
            for j in range(i):
                if check(words[j], words[i]):
                    dp[i] = max(dp[i], dp[j]+1)
        return max(dp)

运行结果

Runtime: 3021 ms, faster than 5.56% of Python online submissions for Longest String Chain.
Memory Usage: 14.2 MB, less than 24.50% of Python online submissions for Longest String Chain.

原题链接

leetcode.com/problems/lo…

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