Leetcode 212. Word Search II

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本文介绍了一种算法，用于在一个二维字符板上查找字典中指定的一系列单词。通过构建字典树并采用深度优先搜索策略，该算法能够高效地找出所有匹配的单词。

Given a 2D board and a list of words from the dictionary, find all words in the board.

Each word must be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.

For example,
Given words = ["oath","pea","eat","rain"] and board =

[
  ['o','a','a','n'],
  ['e','t','a','e'],
  ['i','h','k','r'],
  ['i','f','l','v']
]

Return ["eat","oath"].

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

public class Solution {
    
    public class TrieNode {
        TrieNode[] next;
        Boolean isWord = false;
        public TrieNode() {
            this.next = new TrieNode[26];
        }
    }
    
    public List<String> findWords(char[][] board, String[] words) {
        Set<String> res = new HashSet<>(); 使用一个Set 来除去重复的
        TrieNode root = buildTrie(words);
        int row = board.length, col = board[0].length;
        boolean[][] visited = new boolean [row][col]; visited 来记录
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                dfs(board, i, j, root, res, visited, "");
            }
        }
        return new ArrayList<String>(res);
    }
    
    public void dfs(char[][] board, int x, int y, TrieNode p, Set<String> res, boolean[][] visited, String temp) {
        if (x < 0 || x >= board.length || y < 0 || y >= board[0].length || visited[x][y] == true) return;
        char c = board[x][y];
        if (p.next[ c - 'a'] == null) return; 
        visited[x][y] = true;
        p = p.next[c - 'a'];
        if (p.isWord == true) res.add(temp + board[x][y]);
        
        dfs(board, x + 1, y, p, res, visited, temp + c);
        dfs(board, x - 1, y, p, res, visited, temp + c);
        dfs(board, x, y + 1, p, res, visited, temp + c);
        dfs(board, x, y - 1, p, res, visited, temp + c);
        
        visited[x][y] = false;
    }
    
    private TrieNode buildTrie(String[] words) {
        TrieNode root = new TrieNode();
        for (String word : words) {
            TrieNode p = root;
            for (char c : word.toCharArray()) {
                int index = c - 'a';
                if (p.next[index] == null) {
                    TrieNode temp = new TrieNode();
                    p.next[index] = temp;
                    p = temp;
                }
                else p = p.next[index];
            }
            p.isWord = true;
        }
        return root;
    }
}

public List<String> findWords(char[][] board, String[] words) {
    List<String> res = new ArrayList<>();
    TrieNode root = buildTrie(words);
    for (int i = 0; i < board.length; i++) {
        for (int j = 0; j < board[0].length; j++) {
            dfs (board, i, j, root, res);
        }
    }
    return res;
}

public void dfs(char[][] board, int i, int j, TrieNode p, List<String> res) {
    char c = board[i][j];
    if (c == '#' || p.next[c - 'a'] == null) return;
    p = p.next[c - 'a'];
    if (p.word != null) {   // found one
        res.add(p.word);
        p.word = null;     // de-duplicate
    }

    board[i][j] = '#';
    if (i > 0) dfs(board, i - 1, j ,p, res); 
    if (j > 0) dfs(board, i, j - 1, p, res);
    if (i < board.length - 1) dfs(board, i + 1, j, p, res); 
    if (j < board[0].length - 1) dfs(board, i, j + 1, p, res); 
    board[i][j] = c;
}

public TrieNode buildTrie(String[] words) {
    TrieNode root = new TrieNode();
    for (String w : words) {
        TrieNode p = root;
        for (char c : w.toCharArray()) {
            int i = c - 'a';
            if (p.next[i] == null) p.next[i] = new TrieNode();
            p = p.next[i];
       }
       p.word = w;
    }
    return root;
}

class TrieNode {
    TrieNode[] next = new TrieNode[26];
    String word;
}