力扣labuladong一刷day54天前缀树

一、208. 实现 Trie (前缀树)

题目链接：https://leetcode.cn/problems/implement-trie-prefix-tree/description/?utm_source=LCUS&utm_medium=ip_redirect&utm_campaign=transfer2china
思路：类似于下图就是前缀树，本质是多叉树，只不过表示子节点的数组是通过字母进行索引的，叶子节点有值来表示。

class Trie {

    Node root = null;
    class Node{
        int v = 0;
        Node[] child = new Node[26];
    }
    public Trie() {

    }

    public void insert(String word) {
        if (search(word)) {
            return;
        }
        root = addNode(root, word, 0);
    }

    public boolean search(String word) {
        Node node = getNode(root, word);
        if (node == null || node.v == 0) {
            return false;
        }
        return true;
    }

    public boolean startsWith(String prefix) {
        return getNode(root, prefix) != null;
    }

    Node getNode(Node node, String word) {
        Node p = node;
        for (int i = 0; i < word.length(); i++) {
            if (p == null) {
                return null;
            }
            p = p.child[word.charAt(i)-'a'];
        }
        return p;
    }

    Node addNode(Node node, String word, int i) {
        if (node == null) {
            node = new Node();
        }
        if (i == word.length()) {
            node.v = 1;
            return node;
        }
        int c = word.charAt(i) - 'a';
        node.child[c] = addNode(node.child[c], word, i+1);
        return node;
    }
}

/**
 * Your Trie object will be instantiated and called as such:
 * Trie obj = new Trie();
 * obj.insert(word);
 * boolean param_2 = obj.search(word);
 * boolean param_3 = obj.startsWith(prefix);
 */

二、648. 单词替换

题目链接：https://leetcode.cn/problems/replace-words/description/?utm_source=LCUS&utm_medium=ip_redirect&utm_campaign=transfer2china
思路：和上题一样，也是前缀树的应用，只不过多了一个最短前缀替换。

class Solution {
    public String replaceWords(List<String> dictionary, String sentence) {
        Trie trie = new Trie();
        for (String s : dictionary) {
            trie.insert(s);
        }
        String[] split = sentence.split(" ");
        StringBuilder bf = new StringBuilder();
        for (String s : split) {
            String ms = trie.minSearch(s);
            if ("".equals(ms)) {
                bf.append(s);
            }else {
                bf.append(ms);
            }
            bf.append(" ");
        }
        bf.deleteCharAt(bf.length()-1);
        return bf.toString();
    }

    class Node {
        int v = 0;
        Node[] child = new Node[26];
    }
    class Trie {
        Node root = null;

        void insert(String word) {
            root = addNode(root, word, 0);
        }

        String minSearch(String word) {
            Node p = root;
            for (int i = 0; i < word.length(); i++) {
                if (p == null) return "";
                if (p.v == 1) {
                    return word.substring(0, i);
                }
                p = p.child[word.charAt(i)-'a'];
            }
            if (p != null && p.v == 1) return word;
            return "";
        }

        boolean search(String word) {
            Node node = getNode(word);
            if (node == null || node.v == 0) return false;
            return true;
        }

        Node getNode(String word) {
            Node p = root;
            for (int i = 0; i < word.length(); i++) {
                if (p == null) return null;
                p = p.child[word.charAt(i)-'a'];
            }
            return p;
        }

        Node addNode(Node node, String word, int i) {
            if (node == null) {
                node = new Node();
            }
            if (i == word.length()) {
                node.v = 1;
                return node;
            }
            int c = word.charAt(i)-'a';
            node.child[c] = addNode(node.child[c], word, i+1);
            return node;
        }
    }
}

三、211. 添加与搜索单词 - 数据结构设计

题目链接：https://leetcode.cn/problems/design-add-and-search-words-data-structure/description/
思路：本题还是前缀树，和上一题类似，唯一不同的点是多了一个模式串匹配。

class WordDictionary {
    Node root = null;
    public WordDictionary() {

    }

    public void addWord(String word) {
        root = addNode(root, word, 0);
    }

    public boolean search(String word) {
        return traverse(root, word, 0);
    }

    boolean traverse(Node node, String word, int i) {
        if (node == null) return false;
        if (i == word.length()) {
            return node.v == 1;
        }
        if ('.' == word.charAt(i)) {
            for (int j = 0; j < 26; j++) {
                boolean flag = traverse(node.child[j], word, i + 1);
                if (flag) return flag;
            }
        }else {
            return traverse(node.child[word.charAt(i)-'a'], word, i+1);
        }
        return false;
    }

    Node addNode(Node node, String word, int i) {
        if (node == null) {
            node = new Node();
        }
        if (i == word.length()) {
            node.v = 1;
            return node;
        }
        int c = word.charAt(i)-'a';
        node.child[c] = addNode(node.child[c], word, i+1);
        return node;
    }

}
class Node {
    int v;
    Node[] child = new Node[26];
}

/**
 * Your WordDictionary object will be instantiated and called as such:
 * WordDictionary obj = new WordDictionary();
 * obj.addWord(word);
 * boolean param_2 = obj.search(word);
 */

四、1804. 实现 Trie （前缀树） II

题目链接：https://leetcode.cn/problems/implement-trie-ii-prefix-tree/description/?utm_source=LCUS&utm_medium=ip_redirect&utm_campaign=transfer2china
思路：本题的前缀树多增加了一个功能就是删除节点，删除节点要考虑的就比较多了，到达value的位置要把数量减一，然后在后序位置删除，如果值大于0直接返回就行，不用删除节点，如果值不大于0就需要看该节点的child是否全为null，如果是返回Null就删除了，不是的话保留。

class Trie {
    Node root;
    public Trie() {

    }

    public void insert(String word) {
        root = addNode(root, word, 0);
    }

    public int countWordsEqualTo(String word) {
        Node node = getNode(word);
        if (null == node) return 0;
        return node.v;
    }

    public int countWordsStartingWith(String prefix) {
        Node node = getNode(prefix);
        if (node == null) return 0;
        return traverse(node, 0);
    }

    public void erase(String word) {
        if (getNode(word) == null) return;
        root = deleteNode(root, word, 0);
    }
    Node deleteNode(Node node, String word, int i) {
        if (node == null) return null;
        if (i == word.length()) {
            if (node.v > 0) node.v--;
        }else {
            int c = word.charAt(i)-'a';
            node.child[c] = deleteNode(node.child[c], word, i+1);
        }
        if (node.v > 0) return node;
        for (int j = 0; j < 26; j++) {
            if (node.child[j] != null) {
                return node;
            }
        }
        return null;
    }
    int traverse(Node node, int num) {
        if (node == null) return 0;
        num = node.v;
        for (int i = 0; i < 26; i++) {
            num += traverse(node.child[i], num);
        }
        return num;
    }

    Node addNode(Node node, String word, int i) {
        if (node == null) {
            node = new Node();
        }
        if (i == word.length()) {
            node.v += 1;
            return node;
        }
        int c = word.charAt(i)-'a';
        node.child[c] = addNode(node.child[c], word, i+1);
        return node;
    }

    Node getNode(String word) {
        Node p = root;
        for (int i = 0; i < word.length(); i++) {
            if (p == null) return null;
            p = p.child[word.charAt(i)-'a'];
        }
        return p;
    }
}
class Node{
    int v = 0;
    Node[] child = new Node[26];
}

/**
 * Your Trie object will be instantiated and called as such:
 * Trie obj = new Trie();
 * obj.insert(word);
 * int param_2 = obj.countWordsEqualTo(word);
 * int param_3 = obj.countWordsStartingWith(prefix);
 * obj.erase(word);
 */

五、677. 键值映射

题目链接：https://leetcode.cn/problems/map-sum-pairs/description/?utm_source=LCUS&utm_medium=ip_redirect&utm_campaign=transfer2china
思路：关键点是前缀查询，先获取到前缀的节点，然后广度优先遍历，前序位置记录节点位置。

class MapSum {
    Node root = null;
    public MapSum() {

    }

    public void insert(String key, int val) {
        root = addNode(root, key, val, 0);
    }

    public int sum(String prefix) {
        Node node = getNode(prefix);
        if (node == null) return 0;
        return traverse(node, 0);
    }
    int traverse(Node node, int num) {
        if (node == null) return 0;
        num = node.v;
        for (int i = 0; i < 26; i++) {
            num += traverse(node.child[i], num);
        }
        return num;
    }
    

    Node getNode(String word) {
        Node p = root;
        for (int i = 0; i < word.length(); i++) {
            if (p == null) return null;
            p = p.child[word.charAt(i)-'a'];
        }
        return p;
    }
    Node addNode(Node node, String word, int value, int i) {
        if (node == null) {
            node = new Node();
        }
        if (i == word.length()) {
            node.v = value;
            return node;
        }
        int c = word.charAt(i) - 'a';
        node.child[c] = addNode(node.child[c], word, value, i+1);
        return node;
    }

}
class Node {
    int v = 0;
    Node[] child = new Node[26];
}