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Table of Contents
一、中文版
二、英文版
三、My answer
四、解题报告
一、中文版
给定一个单词列表,我们将这个列表编码成一个索引字符串 S 与一个索引列表 A。
例如,如果这个列表是 ["time", "me", "bell"],我们就可以将其表示为 S = "time#bell#" 和 indexes = [0, 2, 5]。
对于每一个索引,我们可以通过从字符串 S 中索引的位置开始读取字符串,直到 "#" 结束,来恢复我们之前的单词列表。
那么成功对给定单词列表进行编码的最小字符串长度是多少呢?
示例:
输入: words = ["time", "me", "bell"]
输出: 10
说明: S = "time#bell#" , indexes = [0, 2, 5] 。
提示:
1 <= words.length <= 2000
1 <= words[i].length <= 7
每个单词都是小写字母 。
二、英文版
Given a list of words, we may encode it by writing a reference string S and a list of indexes A.
For example, if the list of words is ["time", "me", "bell"], we can write it as S = "time#bell#" and indexes = [0, 2, 5].
Then for each index, we will recover the word by reading from the reference string from that index until we reach a "#" character.
What is the length of the shortest reference string S possible that encodes the given words?
Example:
Input: words = ["time", "me", "bell"]
Output: 10
Explanation: S = "time#bell#" and indexes = [0, 2, 5].
Note:
1 <= words.length <= 2000.
1 <= words[i].length <= 7.
Each word has only lowercase letters.
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/short-encoding-of-words
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
三、My answer
本题采用 LeetCode 的官方题解,记录如下:
class Solution:def minimumLengthEncoding(self, words: List[str]) -> int:good = set(words)for word in words:for k in range(1, len(word)):good.discard(word[k:])return sum(len(word) + 1 for word in good)作者:LeetCode-Solution
链接:https://leetcode-cn.com/problems/short-encoding-of-words/solution/dan-ci-de-ya-suo-bian-ma-by-leetcode-solution/
来源:力扣(LeetCode)
著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。
四、解题报告
记录官方题解是因为代码特别 pythonic,值得学习。
比如 set 的discard() 函数:https://www.runoob.com/python3/ref-set-discard.html
解题思路动图如下:
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