LeetCode-212-单词搜索Ⅱ

题目

给定一个 m x n 二维字符网格 board 和一个单词（字符串）列表 words， 返回所有二维网格上的单词 。

单词必须按照字母顺序，通过 相邻的单元格 内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母在一个单词中不允许被重复使用。

示例 1：

img

输入：board = [["o","a","a","n"],["e","t","a","e"],["i","h","k","r"],["i","f","l","v"]], words = ["oath","pea","eat","rain"]
输出：["eat","oath"]

示例 2：

img

输入：board = [["a","b"],["c","d"]], words = ["abcb"]
输出：[]

提示：

• m == board.length
• n == board[i].length
• 1 <= m, n <= 12
• board[i][j] 是一个小写英文字母
• 1 <= words.length <= 3 * 104
• 1 <= words[i].length <= 10
• words[i] 由小写英文字母组成
• words 中的所有字符串互不相同

题解

class Solution {

    Set<String> set = new HashSet<>();
    char[][] board;
    int n, m;
    TrieNode root = new TrieNode();
    int[][] dirs = new int[][] { { 1, 0 }, { -1, 0 }, { 0, 1 }, { 0, -1 } };
    boolean[][] vis = new boolean[15][15];

    class TrieNode {
        String s;
        TrieNode[] tns = new TrieNode[26];
    }

    void insert(String s) {
        TrieNode p = root;
        for (int i = 0; i < s.length(); i++) {
            int u = s.charAt(i) - 'a';
            if (p.tns[u] == null)
                p.tns[u] = new TrieNode();
            p = p.tns[u];
        }
        p.s = s;
    }

    public List<String> findWords(char[][] _board, String[] words) {
        board = _board;
        m = board.length;
        n = board[0].length;
        for (String w : words)
            insert(w);
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                int u = board[i][j] - 'a';
                if (root.tns[u] != null) {
                    vis[i][j] = true;
                    dfs(i, j, root.tns[u]);
                    vis[i][j] = false;
                }
            }
        }
        List<String> ans = new ArrayList<>();
        for (String s : set)
            ans.add(s);
        return ans;
    }

    void dfs(int i, int j, TrieNode node) {
        if (node.s != null)
            set.add(node.s);
        for (int[] d : dirs) {
            int dx = i + d[0], dy = j + d[1];
            if (dx < 0 || dx >= m || dy < 0 || dy >= n)
                continue;
            if (vis[dx][dy])
                continue;
            int u = board[dx][dy] - 'a';
            if (node.tns[u] != null) {
                vis[dx][dy] = true;
                dfs(dx, dy, node.tns[u]);
                vis[dx][dy] = false;
            }
        }
    }
}