[LeetCode] 286. Walls and Gates
作者:互联网
墙与门。题意是给一个二维矩阵,里面的-1代表墙,0代表门,INF代表一个空的房间。请改写所有的INF,表明每个INF到最近的门的距离。例子
Example:
Given the 2D grid:
INF -1 0 INF INF INF INF -1 INF -1 INF -1 0 -1 INF INFAfter running your function, the 2D grid should be:
3 -1 0 1 2 2 1 -1 1 -1 2 -1 0 -1 3 4
这一题还是flood fill类的题目,既然是问最短距离,所以我这个题目就用BFS做了。
时间O(mn)
空间O(mn)
Java实现
1 class Solution { 2 public void wallsAndGates(int[][] rooms) { 3 // corner case 4 if (rooms == null || rooms.length == 0) { 5 return; 6 } 7 8 // normal case 9 Queue<int[]> queue = new LinkedList<>(); 10 for (int i = 0; i < rooms.length; i++) { 11 for (int j = 0; j < rooms[0].length; j++) { 12 if (rooms[i][j] == 0) { 13 queue.add(new int[] { i, j }); 14 } 15 } 16 } 17 while (!queue.isEmpty()) { 18 int[] cur = queue.poll(); 19 int row = cur[0], col = cur[1]; 20 if (row > 0 && rooms[row - 1][col] == Integer.MAX_VALUE) { 21 rooms[row - 1][col] = rooms[row][col] + 1; 22 queue.offer(new int[] { row - 1, col }); 23 } 24 if (row < rooms.length - 1 && rooms[row + 1][col] == Integer.MAX_VALUE) { 25 rooms[row + 1][col] = rooms[row][col] + 1; 26 queue.offer(new int[] { row + 1, col }); 27 } 28 if (col > 0 && rooms[row][col - 1] == Integer.MAX_VALUE) { 29 rooms[row][col - 1] = rooms[row][col] + 1; 30 queue.offer(new int[] { row, col - 1 }); 31 } 32 if (col < rooms[0].length - 1 && rooms[row][col + 1] == Integer.MAX_VALUE) { 33 rooms[row][col + 1] = rooms[row][col] + 1; 34 queue.offer(new int[] { row, col + 1 }); 35 } 36 } 37 } 38 }
标签:Gates,int,LeetCode,queue,Walls,rooms,INF,col,row 来源: https://www.cnblogs.com/aaronliu1991/p/12817079.html