820. Short Encoding of Words(Leetcode每日一题-2020.03.28)

Problem

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?

Note:

• 1 <= words.length <= 2000.
• 1 <= words[i].length <= 7.
• Each word has only lowercase letters.

Example

Input: words = [“time”, “me”, “bell”]
Output: 10
Explanation: S = “time#bell#” and indexes = [0, 2, 5].

Solution

如果一个单词是另一个单词的后缀，那么它就不用被加入到压缩后的字符串当中。
本题可以使用字典树（Trie）解决，字典树中存储的是倒序的字符串。将输入字符串数组中的每个字符串插入字典树之前，需要先将输入数组按照字符串长度从长倒短进行排序。

注意调用STL的sort对字符串数组按长度排序的方法。

class Solution {
public:

    class TrieNode 
    {
        //char val;
        //TrieNode* children[26];

    public: 
        TrieNode* children[26];
        TrieNode() 
        {
            for(int i = 0;i<26;++i)
            {
                children[i] = NULL;
            }
        }
    };

    // 定义tire
    class Trie 
    {

    
        TrieNode* root;
    public:
        Trie() 
        {
            root = new TrieNode();
        }

        int insert(string &word) 
        {
            TrieNode *cur = root;
            bool isNew = false;
            // 倒着插入单词
            for (int i = word.length() - 1; i >= 0; i--) 
            {
                int c = word[i] - 'a';
                if (cur->children[c] == NULL) 
                {
                    isNew = true; // 是新单词
                    cur->children[c] = new TrieNode();
                }
                cur = cur->children[c];
            }
            // 如果是新单词的话编码长度增加新单词的长度+1，否则不变。
            return isNew? word.length() + 1: 0;
        }
    };

    static bool myfunction (string &i,string &j) 
    { 
        return i.length()>j.length();
    } 

    int minimumLengthEncoding(vector<string>& words) {
        int len = 0;
        Trie *trie = new Trie(); 

        sort(words.begin(),words.end(),myfunction);

        for (auto &word: words) {
            len += trie->insert(word);
        }
        return len;


    }
};