大厂算法面试之 leetcode 精讲 22. 字典树

var Trie = function() {    this.children = {};};
Trie.prototype.insert = function(word) {    let nodes = this.children;    for (const ch of word) {//循环word        if (!nodes[ch]) {//当前字符不在子节点中 则创建一个子节点到children的响应位置            nodes[ch] = {};        }        nodes = nodes[ch];//移动指针到下一个字符子节点    }    nodes.isEnd = true;//字符是否结束};
Trie.prototype.searchPrefix = function(prefix) {    let nodes = this.children;    for (const ch of prefix) {//循环前缀        if (!nodes[ch]) {//当前字符不在子节点中 直接返回false            return false;        }        nodes = nodes[ch];//移动指针到下一个字符子节点    }    return nodes;//返回最后的节点}
Trie.prototype.search = function(word) {    const nodes = this.searchPrefix(word);    //判断searchPrefix返回的节点是不是字符串的结尾的字符    return nodes !== undefined && nodes.isEnd !== undefined;};
Trie.prototype.startsWith = function(prefix) {    return this.searchPrefix(prefix);};

//javaclass Trie {    private Trie[] children;    private boolean isEnd;
    public Trie() {        children = new Trie[26];        isEnd = false;    }        public void insert(String word) {        Trie node = this;        for (int i = 0; i < word.length(); i++) {            char ch = word.charAt(i);            int index = ch - 'a';            if (node.children[index] == null) {                node.children[index] = new Trie();            }            node = node.children[index];        }        node.isEnd = true;    }        public boolean search(String word) {        Trie node = searchPrefix(word);        return node != null && node.isEnd;    }        public boolean startsWith(String prefix) {        return searchPrefix(prefix) != null;    }
    private Trie searchPrefix(String prefix) {        Trie node = this;        for (int i = 0; i < prefix.length(); i++) {            char ch = prefix.charAt(i);            int index = ch - 'a';            if (node.children[index] == null) {                return null;            }            node = node.children[index];        }        return node;    }}

var findWords = function (board, words) {    const trie = new Trie();    const dxys = [        [0, -1],        [-1, 0],        [0, 1],        [1, 0],    ];    const xLen = board.length,        yLen = board[0].length;    const visited = {};    const ret = [];
    // 构建Trie    for (let word of words) {        trie.insert(word);    }
    // DFS board    const dfs = (x, y, nodes, str) => {        if (nodes[board[x][y]].isEnd) {            ret.push(str + board[x][y]);            // 置为false是为了防止重复将字符串加入到ret中            nodes[board[x][y]].isEnd = false;        }
        // 处理本层状态        nodes = nodes[board[x][y]];        str += board[x][y];
        // 向四联通方向检索        visited[x * 100 + y] = true;        for (let [dx, dy] of dxys) {            const newX = x + dx,                newY = y + dy;
            if (                newX < 0 ||                newY < 0 ||                newX >= xLen ||                newY >= yLen ||                !nodes[board[newX][newY]] ||                visited[newX * 100 + newY]            )                continue;
            dfs(newX, newY, nodes, str);        }        visited[x * 100 + y] = false;    };
    for (let x = 0; x < xLen; x++) {        for (let y = 0; y < yLen; y++) {            if (trie.children[board[x][y]]) dfs(x, y, trie.children, "");        }    }
    return ret;};
var Trie = function () {    this.children = {};};
Trie.prototype.insert = function (word) {    let nodes = this.children;    for (const ch of word) {//循环word        if (!nodes[ch]) {//当前字符不在子节点中 则创建一个子节点到children的响应位置            nodes[ch] = {};        }        nodes = nodes[ch];//移动指针到下一个字符    }    nodes.isEnd = true;//字符是否结束};

class Solution {    int[][] dirs = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
    public List<String> findWords(char[][] board, String[] words) {        Trie trie = new Trie();        for (String word : words) {            trie.insert(word);        }
        Set<String> ans = new HashSet<String>();        for (int i = 0; i < board.length; ++i) {            for (int j = 0; j < board[0].length; ++j) {                dfs(board, trie, i, j, ans);            }        }
        return new ArrayList<String>(ans);    }
    public void dfs(char[][] board, Trie now, int i1, int j1, Set<String> ans) {        if (!now.children.containsKey(board[i1][j1])) {            return;        }        char ch = board[i1][j1];        now = now.children.get(ch);        if (!"".equals(now.word)) {            ans.add(now.word);        }
        board[i1][j1] = '#';        for (int[] dir : dirs) {            int i2 = i1 + dir[0], j2 = j1 + dir[1];            if (i2 >= 0 && i2 < board.length && j2 >= 0 && j2 < board[0].length) {                dfs(board, now, i2, j2, ans);            }        }        board[i1][j1] = ch;    }}
class Trie {    String word;    Map<Character, Trie> children;    boolean isWord;
    public Trie() {        this.word = "";        this.children = new HashMap<Character, Trie>();    }
    public void insert(String word) {        Trie cur = this;        for (int i = 0; i < word.length(); ++i) {            char c = word.charAt(i);            if (!cur.children.containsKey(c)) {                cur.children.put(c, new Trie());            }            cur = cur.children.get(c);        }        cur.word = word;    }}

var longestWord = function (words) {    let set = new Set()    words.forEach(v => set.add(v))//set方便查找    //先按长度排序，在按字典序    words.sort((a, b) => a.length === b.length ? a.localeCompare(b) : b.length - a.length)    for (let i = 0; i < words.length; i++) {        let flag = true        for (let j = 1; j < words[i].length; j++) {            if (!set.has(words[i].substring(0, j))) {//查看set中是否有该字符串的每个子串                flag = false                break            }        }        if (flag) {            return words[i]        }    }    return ''};

class Solution {    public String longestWord(String[] words) {        Set<String> wordset = new HashSet();        for (String word: words) wordset.add(word);        Arrays.sort(words, (a, b) -> a.length() == b.length()                    ? a.compareTo(b) : b.length() - a.length());        for (String word: words) {            boolean flag = true;            for (int k = 1; k < word.length(); ++k) {                if (!wordset.contains(word.substring(0, k))) {                    flag = false;                    break;                }            }            if (flag) return word;        }
        return "";    }}

var longestWord = function (words) {    const trie = new Trie()    words.forEach(word => {//将所有字符串插入trie中        trie.insert(word)    })    let res = ''    const _helper = (nodes, path) => {        if (path.length > res.length || (res.length === path.length && res > path)) {            res = path        }        //{a:{b1:{c1:{isEnd: true}},b2:{c2:{isEnd: true}}}}        for (const [key, value] of Object.entries(nodes)) {                    if (value.isEnd) {//如果当前字符是某一个字符串的结尾                path += key                _helper(value, path)//递归寻找                path = path.slice(0, -1)//回溯            }        }    }    _helper(trie.children, '')//递归寻找那个长度最大的单词    return res}
var Trie = function() {    this.children = {};};
Trie.prototype.insert = function(word) {    let nodes = this.children;    for (const ch of word) {//循环word        if (!nodes[ch]) {//当前字符不在子节点中 则创建一个子节点到children的响应位置            nodes[ch] = {};        }        nodes = nodes[ch];//移动指针到下一个字符    }    nodes.isEnd = true;//字符是否结束};