字母迷宫

/*
 * 思路：
 * 1. 使用Stream API来简化遍历操作。
 * 2. 由于Stream API不适合处理复杂的递归操作，因此这里只能在基础遍历中使用。
 * 3. 核心逻辑仍然使用DFS进行深度优先搜索。
 */

import java.util.stream.IntStream;

public class Solution {
    public boolean exist(char[][] board, String word) {
        int m = board.length;
        int n = board[0].length;
        boolean[][] visited = new boolean[m][n];
        return IntStream.range(0, m)
                .boxed()
                .flatMap(i -> IntStream.range(0, n)
                        .mapToObj(j -> new int[]{i, j}))
                .anyMatch(pos -> dfs(board, word, 0, pos[0], pos[1], visited));
    }

    private boolean dfs(char[][] board, String word, int index, int x, int y, boolean[][] visited) {
        if (index == word.length()) {
            return true;
        }
        if (x < 0 || y < 0 || x >= board.length || y >= board[0].length || visited[x][y] || board[x][y] != word.charAt(index)) {
            return false;
        }
        visited[x][y] = true;
        boolean found = dfs(board, word, index + 1, x + 1, y, visited) ||
                        dfs(board, word, index + 1, x - 1, y, visited) ||
                        dfs(board, word, index + 1, x, y + 1, visited) ||
                        dfs(board, word, index + 1, x, y - 1, visited);
        visited[x][y] = false;
        return found;
    }
}