286. Walls and Gates
Problem:
You are given a m x n 2D grid initialized with these three possible values.
-1- A wall or an obstacle.0- A gate.INF- Infinity means an empty room. We use the value231 - 1 = 2147483647to representINFas you may assume that the distance to a gate is less than2147483647.
Fill each empty room with the distance to its nearest gate. If it is impossible to reach a gate, it should be filled with
INF.
For example, given the 2D grid:
INF -1 0 INF INF INF INF -1 INF -1 INF -1 0 -1 INF INF
After running your function, the 2D grid should be:
3 -1 0 1 2 2 1 -1 1 -1 2 -1 0 -1 3 4
Analysis:
Standard reverse BFS. 01 Matrix 的马甲题。
Solution:
class Solution { public void wallsAndGates(int[][] rooms) { if (rooms == null || rooms.length == 0 || rooms[0].length == 0) return; int m = rooms.length; int n = rooms[0].length; int[][] dirs = new int[][]{{0, 1}, {0, -1}, {1, 0}, {-1, 0}}; Queue<int[]> queue = new LinkedList<>(); for (int i = 0; i < m; i++) for (int j = 0; j < n; j++) { if (rooms[i][j] == 0) { queue.offer(new int[]{i, j}); } } int dis = 0; while (!queue.isEmpty()) { int size = queue.size(); dis++; for (int i = 0; i < size; i++) { int[] cur = queue.poll(); int x = cur[0]; int y = cur[1]; for (int[] dir: dirs) { int newx = x + dir[0]; int newy = y + dir[1]; int newIndex = newx * n + newy; if (newx < 0 || newx >= m || newy < 0 || newy >= n) continue; if (rooms[newx][newy] != 2147483647 ) continue; queue.offer(new int[]{newx, newy}); rooms[newx][newy] = dis; } } } } }
评论
发表评论