208. Implement Trie (Prefix Tree)-优快云博客

Implement a trie with insert, search, and startsWith methods.

Note:

You may assume that all inputs are consist of lowercase letters a-z.

分清trie和trieNode

第一次做的时候感觉每一个trieNode都可以作为trie 形成了递归的关系写起来就比较乱

solution:

class TrieNode {

    // R links to node children
    private TrieNode[] links;

    private final int R = 26;

    private boolean isEnd;

    public TrieNode() {
        links = new TrieNode[R];
    }

    public boolean containsKey(char ch) {
        return links[ch -'a'] != null;
    }
    public TrieNode get(char ch) {
        return links[ch -'a'];
    }
    public void put(char ch, TrieNode node) {
        links[ch -'a'] = node;
    }
    public void setEnd() {
        isEnd = true;
    }
    public boolean isEnd() {
        return isEnd;
    }
}


class Trie {
    private TrieNode root;

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

    // Inserts a word into the trie.
    public void insert(String word) {
        TrieNode node = root;
        for (int i = 0; i < word.length(); i++) {
            char currentChar = word.charAt(i);
            if (!node.containsKey(currentChar)) {
                node.put(currentChar, new TrieNode());
            }
            node = node.get(currentChar);
        }
        node.setEnd();
    }


    // search a prefix or whole key in trie and
    // returns the node where search ends
    private TrieNode searchPrefix(String word) {
        TrieNode node = root;
        for (int i = 0; i < word.length(); i++) {
           char curLetter = word.charAt(i);
           if (node.containsKey(curLetter)) {
               node = node.get(curLetter);
           } else {
               return null;
           }
        }
        return node;
    }

    // Returns if the word is in the trie.
    public boolean search(String word) {
       TrieNode node = searchPrefix(word);
       return node != null && node.isEnd();
    }

     // Returns if there is any word in the trie
    // that starts with the given prefix.
    public boolean startsWith(String prefix) {
        TrieNode node = searchPrefix(prefix);
        return node != null;
    }
}

自己的和solution还有一点主要的区别对于子节点 solution用的是char数组我用的是set

时间复杂度是相同的但空间复杂度上显然我的会比较高

因为hashset实际上是封装的hashmap hashmap底层是链表的数组为避免hash冲突会开辟比较大的空间

对于这点题目其实也有提示

You may assume that all inputs are consist of lowercase letters a-z.

所以当需要map或者set 但是元素只有字母的时候你可以使用数组来代替

比如先要统计一个非常长的单词每个字母出现的次数首先会想到hashmap

但实际上使用一个长度为26的int数组就可以了

int[] map = new int[26];

map[0]表示a出现的次数 map[1]表示b出现的次数。。。

获取索引及对应的值也非常方便

比如 f出现的次数对应的值索引就是 ’f’-‘a’ = 5 出现的次数就是map[5]