Given an N x N grid containing only values 0 and 1, where 0 represents water and 1 represents land, find a water cell such that its distance to the nearest land cell is maximized and return the distance.
The distance used in this problem is the Manhattan distance: the distance between two cells (x0, y0) and (x1, y1) is |x0 - x1| + |y0 - y1|.
If no land or water exists in the grid, return -1.
Example 1:
Input: [[1,0,1],[0,0,0],[1,0,1]]
Output: 2
Explanation:
The cell (1, 1) is as far as possible from all the land with distance 2.
Example 2:
Input: [[1,0,0],[0,0,0],[0,0,0]]
Output: 4
Explanation:
The cell (2, 2) is as far as possible from all the land with distance 4.
Note:
1 <= grid.length == grid[0].length <= 100
grid[i][j] is 0 or 1
classSolution {int[][] directions =newint[][] { { -1,0 }, { 1,0 }, { 0,-1 }, { 0,1 } };publicintmaxDistance(int[][] grid) {if (grid ==null||grid.length==0|| grid[0].length==0)return-1;boolean[][] visited =newboolean[grid.length][grid[0].length];Queue<int[]> queue =newArrayDeque<>();for (int i =0; i <grid.length; i++) {for (int j =0; j < grid[0].length; j++) {if (grid[i][j] ==1) {queue.add(newint[] { i, j }); visited[i][j] =true; } } }int level =-1;while (!queue.isEmpty()) {int size =queue.size();for (int i =0; i < size; i++) {int[] start =queue.poll();int x = start[0];int y = start[1];for (int[] dir : directions) {int newX = x + dir[0];int newY = y + dir[1]; if (newX >= 0 && newX < grid.length && newY >= 0 && newY < grid[0].length && !visited[newX][newY]) {
visited[newX][newY] =true;queue.add(newint[] { newX, newY }); } } } level++; }// If no land or water exists in the grid, return -1return level <= 0 ? -1 : level; }}