LeetCode [1091] Shortest Path in Binary Matrix

1091. Shortest Path in Binary Matrix

Medium

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.

 

Example 1:

Input: [[0,1],[1,0]]


Output: 2

Example 2:

Input: [[0,0,0],[1,1,0],[1,1,0]]


Output: 4

 

Note:

1. 1 <= grid.length == grid[0].length <= 100
2. grid[r][c] is 0 or 1

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class Solution { public int shortestPathBinaryMatrix(int[][] grid) { int[][] dirs = new int[][] { { -1, 0 }, { 1, 0 }, { 0, -1 }, { 0, 1 }, { 1, 1 }, { 1, -1 }, { -1, 1 }, { -1, -1 } }; int m = grid.length, n = grid[0].length; if (grid[0][0] != 0) return -1; Queue<int[]> que = new LinkedList<>(); que.add(new int[] { 0, 0 }); grid[0][0] = 1; int steps = 1; while (!que.isEmpty()) { int sz = que.size(); for (int i = 0; i < sz; ++i) { int[] top = que.poll(); int x = top[0], y = top[1]; if (x == m - 1 && y == n - 1) return steps; for (int[] d : dirs) { int x1 = x + d[0], y1 = y + d[1]; if (x1 >= 0 && x1 < m && y1 >= 0 && y1 < n && grid[x1][y1] == 0) { que.add(new int[] { x1, y1 }); grid[x1][y1] = 1; } } } steps++; } return -1; } }