1. 程式人生 > 實用技巧 >Leetcode 211 新增與搜尋單詞 字首樹

Leetcode 211 新增與搜尋單詞 字首樹

  JAVA:

class WordDictionary {
        private Node head;

        /**
         * Initialize your data structure here.
         */
        public WordDictionary() {
            this.head = new Node(null);
        }

        /**
         * Adds a word into the data structure.
         */
        public
void addWord(String word) { Node node = this.head; for (int i = 0; i < word.length(); i++) { int chPoint = word.charAt(i) - 'a'; if (node.childs[chPoint] == null) node.childs[chPoint] = new Node(node); else node.childs[chPoint].nums++; node.hasChild
= true; node = node.childs[chPoint]; } node.isEnd = true; node.pre.hasEndChild = true; } public boolean search(String word) { return search(word, this.head); } /** * Returns if the word is in the data structure. A word could contain the dot character '.' to represent any * one letter.
*/ public boolean search(String word, Node node) { int len = word.length(); for (int i = 0; i < len; i++) { // 包含 . 進行全匹配 if (word.charAt(i) == '.') { if (i == len - 1) return node.hasEndChild; for (int j = 0; j < 26; j++) { if (node.childs[j] == null) continue; if (search(word.substring(i + 1, len), node.childs[j])) return true; } return false; } // 正常匹配 int chPoint = word.charAt(i) - 'a'; if (node.childs[chPoint] == null) return false; node = node.childs[chPoint]; } return node.isEnd; } private class Node { boolean isEnd; boolean hasChild; boolean hasEndChild; Node pre; Node[] childs; int nums; Node(Node pre) { this.isEnd = false; this.childs = new Node[26]; this.hasChild = false; this.nums = 1; this.pre = pre; this.hasEndChild = false; } } }

  JS:

/**
 * Initialize your data structure here.
 */
var WordDictionary = function () {
    this.head = new Node(null);
};

/**
 * Adds a word into the data structure.
 * @param {string} word
 * @return {void}
 */
WordDictionary.prototype.addWord = function (word) {
    Node
    node = this.head;
    for (let i = 0; i < word.length; i++) {
        let ch = word.charAt(i), chPoint = ch.charCodeAt() - 97;
        if (!node.childs[chPoint]) node.childs[chPoint] = new Node(node);
        else node.childs[chPoint].nums++;
        node = node.childs[chPoint];
    }
    node.isEnd = true;
    node.pre.hasEndChild = true;
};

/**
 * Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter.
 * @param {string} word
 * @return {boolean}
 */
WordDictionary.prototype.search = function (word) {
    return this.search0(word, this.head);
}


WordDictionary.prototype.search0 = function (word, node) {
    let len = word.length;
    for (let i = 0; i < len; i++) {
        let ch = word.charAt(i), chPoint = ch.charCodeAt() - 97;
        if (ch == '.') {
            if (i == len - 1) return node.hasEndChild;
            for (let j = 0; j < 26; j++) {
                if (!node.childs[j]) continue;
                if (this.search0(word.substring(i + 1, len), node.childs[j])) return true;
            }
            return false;
        }
        if (!node.childs[chPoint]) return false;
        node = node.childs[chPoint];
    }
    return node.isEnd;
};

var Node = function (pre) {
    this.isEnd = false;
    this.childs = new Array(26);
    this.hasEndChild = false;
    this.pre = pre;
    this.nums = 1;
}

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