Java 实现二进制矩阵最短畅通路径算法

思路：BFS

由于要求最短路径，可以使用 BFS 来解决。从左上角出发，每次遍历相邻的 8 个方向，如果该方向上的格子可以通行且未被访问过，则加入队列中，并标记为已访问。每次遍历时，记录下当前遍历的层数，即路径长度，直到到达右下角或者队列为空为止。

代码如下：

public class Solution {
    public int shortestPathBinaryMatrix(int[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0 || grid[0][0] == 1 || grid[grid.length - 1][grid[0].length - 1] == 1) {
            return -1;
        }
        int[][] directions = {{-1, -1}, {-1, 0}, {-1, 1}, {0, -1}, {0, 1}, {1, -1}, {1, 0}, {1, 1}};
        int n = grid.length;
        Queue<int[]> queue = new LinkedList<>();
        queue.offer(new int[]{0, 0});
        boolean[][] visited = new boolean[n][n];
        visited[0][0] = true;
        int pathLength = 1;
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                int[] current = queue.poll();
                int row = current[0];
                int col = current[1];
                if (row == n - 1 && col == n - 1) {
                    return pathLength;
                }
                for (int[] direction : directions) {
                    int nextRow = row + direction[0];
                    int nextCol = col + direction[1];
                    if (nextRow >= 0 && nextRow < n && nextCol >= 0 && nextCol < n && grid[nextRow][nextCol] == 0 && !visited[nextRow][nextCol]) {
                        queue.offer(new int[]{nextRow, nextCol});
                        visited[nextRow][nextCol] = true;
                    }
                }
            }
            pathLength++;
        }
        return -1;
    }
}