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：
求矩阵中room到gate的最短距离。这里我们可以用BFS， 先把所有gate放入queue中，再根据gate计算上下左右，假如有room,再把这个room加入到queue中。
Time Complexity - O(4n)， Space Complexity - O(4n)
public class Solution { public void wallsAndGates(int[][] rooms) { if(rooms == null || rooms.length == 0) { return; } Queue<int[]> queue = new LinkedList<>(); for(int i = 0; i < rooms.length; i++) { for(int j = 0; j < rooms[0].length; j++) { if(rooms[i][j] == 0) { queue.add(new int[]{i, j}); } } } while(!queue.isEmpty()) { int[] gate = queue.poll(); int row = gate[0], col = gate[1]; if(row > 0 && rooms[row - 1][col] == Integer.MAX_VALUE) { rooms[row - 1][col] = rooms[row][col] + 1; queue.add(new int[]{row - 1, col}); } if(row < rooms.length - 1 && rooms[row + 1][col] == Integer.MAX_VALUE) { rooms[row + 1][col] = rooms[row][col] + 1; queue.add(new int[]{row + 1, col}); } if(col > 0 && rooms[row][col - 1] == Integer.MAX_VALUE) { rooms[row][col - 1] = rooms[row][col] + 1; queue.add(new int[]{row, col - 1}); } if(col < rooms[0].length - 1 && rooms[row][col + 1] == Integer.MAX_VALUE) { rooms[row][col + 1] = rooms[row][col] + 1; queue.add(new int[]{row, col + 1}); } } } }
Reference:
https://www.cs.ubc.ca/~kevinlb/teaching/cs322%20-%202008-9/Lectures/Search3.pdf
https://en.wikipedia.org/wiki/Breadth-first_search#Time_and_space_complexity
https://leetcode.com/discuss/60552/beautiful-java-solution-10-lines
https://leetcode.com/discuss/60418/c-bfs-clean-solution-with-simple-explanations
https://leetcode.com/discuss/60170/6-lines-o-mn-python-bfs
https://leetcode.com/discuss/73686/concise-java-solution-bfs-7ms
https://leetcode.com/discuss/68456/java-dfs-solution-much-quicker-and-simpler-than-bfs
原文：http://www.cnblogs.com/yrbbest/p/5038632.html