[Leedcode] [JAVA] [820 title] [trie] [Set]

【Problem Description】

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 。

[Thinking] answer

1. Add set in the array, each word in the suffix cutting set, remove the same suffix (e.g., cutting time excluding me)

• word.substring (n) -> from the n subscripts begin cutting (n <word.length ())
  Examples
• time.substring(1) -> ime
• time.substring(2) -> me
• time.substring(3) -> e

class Solution {
    public int minimumLengthEncoding(String[] words) {
        Set<String> set = new HashSet<>(Arrays.asList(words));
        for (String word : words) {
            for (int i = 1; i < word.length(); i++) {
                set.remove(word.substring(i));
            }
        }
        int ans = 0;
//+1  按照题意#
        for (String word : set) {
            ans += word.length() + 1;
        }
        return ans;
    }
}

2. trie / Trie tree / prefix tree O (N ^ 2)

• Reverse insertion of the word trie longer (suffix) length Override
• Trie judge whether there is a word in the dictionary in reverse tree
  

class Solution {
    public int minimumLengthEncoding(String[] words) {
        int len = 0;
        Trie trie = new Trie();
        // 先对单词列表根据单词长度由长到短排序
        Arrays.sort(words, (s1, s2) -> s2.length() - s1.length());
        // 单词插入trie，返回该单词增加的编码长度
        for (String word: words) {
            len += trie.insert(word);
        }
        return len;
    }
}

// 定义tire
class Trie {
    
    TrieNode root;
    
    public Trie() {
        root = new TrieNode();
    }

    public int insert(String word) {
        TrieNode cur = root;
        boolean isNew = false;
        // 倒着插入单词
        for (int i = word.length() - 1; i >= 0; i--) {
            int c = word.charAt(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;
    }
}

class TrieNode {
    char val;
    TrieNode[] children = new TrieNode[26];

    public TrieNode() {}
}

作者：sweetiee
链接：https://leetcode-cn.com/problems/short-encoding-of-words/solution/99-java-trie-tu-xie-gong-lue-bao-jiao-bao-hui-by-s/

【to sum up】

1. length property length () method
   -java is not an array length () method, only the length property, the array is returned array.length length of the array.

- String String is length () method, str.length () returns the length of the string.

2. trie appears place

• Lenovo search engine field
• Square block chain Ethernet Merkle Patricia Tree Merkel prefix tree tree +
• English word