Create 0286.(Medium) Walls and Gates.java

Author: haibarawu

Algorithms/0286.(Medium) Walls and Gates.java

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+/****************************************************************************************************
+286. Walls and Gates
+
+Difficulty: Medium
+
+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.
+
+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
+****************************************************************************************************/
+
+
+class Solution {
+    public void wallsAndGates(int[][] rooms) {
+        //记录四个方向，向 上下左右（i +- 1, j +- 1）方向前进。
+        final int[][] DIRS = new int[][] {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
+        // if (rooms == null || rooms.length == 0 || rooms[0].length == 0) {
+        //     return rooms;
+        // }
+        
+        Queue<int[]> queue = new LinkedList<>();
+        //遍历一遍二维数组，将为 0 （即是 gate 的格子）放入队列 queue 中。
+        for (int i = 0; i < rooms.length; i++) {
+            for (int j = 0; j < rooms[0].length; j++) {
+                if (rooms[i][j] == 0) {
+                    queue.offer(new int[] {i, j});
+                }
+            }
+        }
+        
+        while (!queue.isEmpty()) {
+            int[] cur = queue.poll();
+            int i = cur[0];
+            int j = cur[1];
+            //从当前点向 上下左右 前进一格。如果前进到达的新的这一格为 INF，则 ＋ 1。并且放入队列 queue 中，未来继续遍历它的四个方向。
+            for (int[] pairs : DIRS) {
+                int x = i + pairs[0];
+                int y = j + pairs[1];
+                if (x < 0 || x >= rooms.length || y < 0 || y >= rooms[0].length || rooms[x][y] != Integer.MAX_VALUE) {
+                    continue;
+                }
+                rooms[x][y] = rooms[i][j] + 1;
+                queue.add(new int[] {x, y});
+            }
+        }
+        //return rooms;
+    }
+}
+
+