1091. Shortest Path in Binary Matrix

In an N by N square grid, each cell is either empty (0) or blocked (1).

A clear path from top-left to bottom-right has length k if and only if it is composed of cells C_1, C_2, ..., C_k such that:

• Adjacent cells C_i and C_{i+1} are connected 8-directionally (ie., they are different and share an edge or corner)
• C_1 is at location (0, 0) (ie. has value grid[0][0])
• C_k is at location (N-1, N-1) (ie. has value grid[N-1][N-1])
• If C_i is located at (r, c), then grid[r][c] is empty (ie. grid[r][c] == 0).

Return the length of the shortest such clear path from top-left to bottom-right.  If such a path does not exist, return -1.

 

class Solution {
    public int shortestPathBinaryMatrix(int[][] grid) {
        int n = grid.length;
        if (grid[0][0] == 1 || grid[n - 1][n - 1] == 1) return -1;

        boolean[][] visited = new boolean[n][n];
        Queue<int[]> q = new LinkedList<>();

        q.offer(new int[] { 0, 0 });
        visited[0][0] = true;
        int[][] directions = { { 0, 1 }, { 0, -1 }, { 1, 0 }, { 1, -1 }, { 1, 1 }, { -1, 0 }, { -1, -1 }, { -1, 1 } };
        int cost = 1;
        while (!q.isEmpty()) {
            int size = q.size();
            for (int i = 0; i < size; i++) {
                int[] cur = q.poll();
                if (cur[0] == n - 1 && cur[1] == n - 1) return cost;
                for (int[] dir : directions) {
                    int x = cur[0] + dir[0];
                    int y = cur[1] + dir[1];
                    if (x >= 0 && x < n && y >= 0 && y < n && !visited[x][y] && grid[x][y] == 0) {
                        q.offer(new int[] { x, y });
                        visited[x][y] = true;
                    }
                }
            }
            cost++;
        }
        return -1;
    }
}