提取咒文

/*
 * 思路：
 * 使用 Java Stream 进行 BFS 搜索每个字母的最短路径。
 * 在 Java Stream 中对队列进行并行处理。
 */
import java.util.*;
import java.util.stream.*;

public class Solution {
    public int minSteps(String[] matrix, String mantra) {
        int rows = matrix.length;
        int cols = matrix[0].length();
        int[][] directions = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};

        class Node {
            int x, y, steps;
            Node(int x, int y, int steps) {
                this.x = x;
                this.y = y;
                this.steps = steps;
            }
        }

        int totalSteps = 0;
        int startX = 0, startY = 0;

        for (char c : mantra.toCharArray()) {
            Queue<Node> queue = new LinkedList<>();
            boolean[][] visited = new boolean[rows][cols];
            queue.add(new Node(startX, startY, 0));
            visited[startX][startY] = true;
            boolean found = false;

            while (!queue.isEmpty()) {
                List<Node> nodeList = new LinkedList<>();
                queue.forEach(nodeList::add);
                queue.clear();

                nodeList.parallelStream().forEach(node -> {
                    if (matrix[node.x].charAt(node.y) == c) {
                        totalSteps += node.steps;
                        startX = node.x;
                        startY = node.y;
                        found = true;
                        queue.clear();
                    }
                    for (int[] dir : directions) {
                        int newX = node.x + dir[0];
                        int newY = node.y + dir[1];
                        if (newX >= 0 && newX < rows && newY >= 0 && newY < cols && !visited[newX][newY]) {
                            queue.add(new Node(newX, newY, node.steps + 1));
                            visited[newX][newY] = true;
                        }
                    }
                });

                if (found) break;
            }

            if (!found) {
                return -1;
            }
        }
        return totalSteps + mantra.length();
    }
}