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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 than 2147483647.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
public class Solution { int[][] moves = new int[][]{{0,1},{0,-1},{1,0},{-1,0}}; public void wallsAndGates(int[][] rooms) { if (rooms.length==0 || rooms[0].length==0) return; for (int i=0;i<rooms.length;i++) for (int j=0;j<rooms[0].length;j++) if (rooms[i][j]==0){ updateDistanceFromGate(rooms,i,j); } } // IMPORTANT: No need to use visited matrix. if a new node has larger distance than the current distance, we do not put this node into queue, as starting from this node we can not get any better result for its neighbors. This makes us being able to avoid visited matrix. public void updateDistanceFromGate(int[][] rooms, int x, int y){ Queue<int[]> queue = new LinkedList<int[]>(); int[] cur = new int[]{x,y,0}; queue.add(cur); while (!queue.isEmpty()){ cur = queue.poll(); for (int i=0;i<4;i++){ int[] next = new int[]{cur[0]+moves[i][0],cur[1]+moves[i][1],cur[2]+1}; if (isValid(rooms,next)) { if (rooms[next[0]][next[1]]>next[2]){ rooms[next[0]][next[1]] = next[2]; queue.add(next); } } } } } public boolean isValid(int[][] rooms, int[] room){ if (room[0]<0 || room[0]>=rooms.length || room[1]<0 || room[1]>=rooms[0].length){ return false; } return (rooms[room[0]][room[1]]>0); } }
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原文地址：http://www.cnblogs.com/lishiblog/p/5855178.html
