refactor 407

fishercoder1534 · fishercoder1534 · commit bb51126926ef · 2020-07-06T06:50:30.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_407.java b/src/main/java/com/fishercoder/solutions/_407.java
@@ -2,33 +2,11 @@
 
 import java.util.PriorityQueue;
 
-/**
- * 407. Trapping Rain Water II
- *
- * Given an m x n matrix of positive integers representing the height of each unit cell in a 2D elevation map, compute the volume of water it is able to trap after raining.
-
- Note:
- Both m and n are less than 110. The height of each unit cell is greater than 0 and is less than 20,000.
-
- Example:
-
- Given the following 3x6 height map:
- [
- [1,4,3,1,3,2],
- [3,2,1,3,2,4],
- [2,3,3,2,3,1]
- ]
-
- Return 4.
-
- The above image represents the elevation map [[1,4,3,1,3,2],[3,2,1,3,2,4],[2,3,3,2,3,1]] before the rain.
-
- After the rain, water are trapped between the blocks. The total volume of water trapped is 4.
-
- */
 public class _407 {
     public static class Solution1 {
-        /** Reference: https://discuss.leetcode.com/topic/60418/java-solution-using-priorityqueue */
+        /**
+         * Reference: https://discuss.leetcode.com/topic/60418/java-solution-using-priorityqueue
+         */
         public class Cell {
             int row;
             int col;
@@ -70,7 +48,7 @@ public int trapRainWater(int[][] heights) {
             // from the borders, pick the shortest cell visited and check its neighbors:
             // if the neighbor is shorter, collect the water it can trap and update its height as its height plus the water trapped
             // add all its neighbors to the queue.
-            int[][] dirs = new int[][] {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
+            int[][] dirs = new int[][]{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
             int res = 0;
             while (!queue.isEmpty()) {
                 Cell cell = queue.poll();
