Tire数据结构

Word Search II

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.
Example:
Input:
words = ["oath","pea","eat","rain"] and board =
[
['o','a','a','n'],
['e','t','a','e'],
['i','h','k','r'],
['i','f','l','v']
]
Output: ["eat","oath"]

public class Solution {
    Set<String> res = new HashSet<String>();
    
    public List<String> findWords(char[][] board, String[] words) {
        Trie trie = new Trie();
        for (String word : words) {
            trie.insert(word);
        }
        
        int m = board.length;
        int n = board[0].length;
        boolean[][] visited = new boolean[m][n];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                dfs(board, visited, "", i, j, trie);
            }
        }
        
        return new ArrayList<String>(res);
    }
    
    public void dfs(char[][] board, boolean[][] visited, String str, int x, int y, Trie trie) {
        if (x < 0 || x >= board.length || y < 0 || y >= board[0].length) return;
        if (visited[x][y]) return;
        
        str += board[x][y];
        if (!trie.startsWith(str)) return;
        
        if (trie.search(str)) {
            res.add(str);
        }
        
        visited[x][y] = true;
        dfs(board, visited, str, x - 1, y, trie);
        dfs(board, visited, str, x + 1, y, trie);
        dfs(board, visited, str, x, y - 1, trie);
        dfs(board, visited, str, x, y + 1, trie);
        visited[x][y] = false;
    }
}

注意：Trie数据结构，字典树，典型应用是用于统计和排序大量的字符串（但不仅限于字符串），所以经常被搜索引擎系统用于文本词频统计。它的优点是：最大限度地减少无谓的字符串比较，查询效率比哈希表高。