LeetCode 286. Walls and Gates

原题链接在这里：https://leetcode.com/problems/walls-and-gates/

题目：

You are given a m x n 2D grid initialized with these three possible values.

  1. -1 - A wall or an obstacle.
  2. 0 - A gate.
  3. INF - Infinity means an empty room. We use the value 231 - 1 = 2147483647 to represent INF as 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

题解：

BFS, 先把所有gate加到que中。对于每一个从que中poll出来的gate，看四个方向是否有room, 若有，把room的值改正gate + 1, 在放回到que中.

Time Complexity: O(m*n). m = rooms.length, n = rooms[0].length. 每个点没有扫描超过两遍. Space: O(m*n). 

AC Java:

 1 class Solution {
 2     public void wallsAndGates(int[][] rooms) {
 3         if(rooms == null || rooms.length == 0 || rooms[0].length == 0){
 4             return;
 5         }
 6         
 7         int m = rooms.length;
 8         int n = rooms[0].length;
 9         LinkedList<int []> que = new LinkedList<>();
10         int level = 1;
11         for(int i = 0; i < m; i++){
12             for(int j = 0; j < n; j++){
13                 if(rooms[i][j] == 0){
14                     que.add(new int[]{i, j});
15                 }
16             }
17         }
18         
19         int [][] dirs = new int[][]{{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
20         
21         while(!que.isEmpty()){
22             int size = que.size();
23             while(size-- > 0){
24                 int [] cur = que.poll();
25                 for(int [] dir : dirs){
26                     int x = cur[0] + dir[0];
27                     int y = cur[1] + dir[1];
28                     if(x < 0 || x >= m || y < 0 || y >= n || rooms[x][y] != Integer.MAX_VALUE){
29                         continue;
30                     }
31                     
32                     que.add(new int[]{x, y});
33                     rooms[x][y] = level;
34                 }        
35             }
36             
37             level++;
38         }
39     }
40 }

跟上Robot Room Cleaner, Rotting Oranges, Shortest Distance from All Buildings.

posted @ 2016-02-29 03:31  Dylan_Java_NYC  阅读(532)  评论(0)  编辑  收藏  举报