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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 - A wall or an obstacle.0 - A gate.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 public class Solution { 2 public void wallsAndGates(int[][] rooms) { 3 if(rooms == null || rooms.length == 0 || rooms[0].length == 0){ 4 return; 5 } 6 int [][] fourDire = {{0,1},{0,-1},{1,0},{-1,0}}; 7 LinkedList<int []> que = new LinkedList<int []>(); 8 9 for(int i = 0; i<rooms.length; i++){ 10 for(int j = 0; j<rooms[0].length; j++){ 11 if(rooms[i][j] == 0){ 12 que.add(new int[]{i,j}); 13 } 14 } 15 } 16 17 while(!que.isEmpty()){ 18 int [] gate = que.poll(); 19 int x = gate[0]; 20 int y = gate[1]; 21 for(int k = 0; k<4; k++){ 22 int newX = x + fourDire[k][0]; 23 int newY = y + fourDire[k][1]; 24 if(newX>=0 && newX<rooms.length && newY>=0 && newY<rooms[0].length){ 25 if(rooms[newX][newY] == Integer.MAX_VALUE){ 26 rooms[newX][newY] = rooms[x][y] + 1; 27 que.add(new int[]{newX, newY}); 28 } 29 } 30 } 31 } 32 } 33 }
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原文地址：http://www.cnblogs.com/Dylan-Java-NYC/p/5226225.html
