[Leetcode] 820. Short Encoding of Words 解题报告_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。-优快云博客

本文链接：https://blog.youkuaiyun.com/magicbean2/article/details/80340096

本文介绍了一种高效算法，用于寻找给定单词列表的最短参考字符串编码方式。通过使用哈希表并删除所有可能的后缀，确保了每个单词及其后缀只被编码一次。

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题目：

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.

思路：

在网上参考了一个非常妙的算法，现在分享如下：

我们将words中的所有字符串都放在一个哈希表中，然后对于哈希表中的每个单词，在哈希表中删除掉它的所有后缀（因为它的后缀可以由它本身encode得到）。最后再计算出来哈希表中所有单词的长度和即可。

代码：

class Solution {
public:
    int minimumLengthEncoding(vector<string>& words) {
        unordered_set<string> hash(words.begin(), words.end());
        for (string w : hash)
            for (int i = 1; i < w.size(); ++i)  // try to erase from the hash set
                hash.erase(w.substr(i));
        int res = 0;
        for (string w : hash) {
            res += w.size() + 1;
        }
        return res;
    }
};