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.
思路：与2D的管住两边的最低高度，3D只要管住包围的环中的最低高度就可以，保证最外围的单元一定在优先队列中，然后每次先遍历原先队列中高度最小的单元A，如果A的四周有尚未被遍历并且高度小于A的单元B，那么A的高度-B的高度就是新增加的蓄水体积，然后设置B的高度=max（A的高度，B的高度），将B加入优先队列。
另外一个需要注意的是解法中对类的比较的定义，Cell执行Comparable借口，重定义compareTo（Cell o）函数：返回负数意味着当前元素“小于”o，返回正数意味着当前元素“大于”o，返回0意味着当前元素“等于”o。
compareTo函数中定义成：
return this.height - o.height;
这就意味着this.height > o.height时，Comparable返回的是当前元素“大于”o；this.height < o.height时，Comparable返回的是当前元素“小于”o。又因为JAVA的PriorityQueue是小顶堆，因此队列的Cell是按height升序排列的。
那么，如果想要队列中的Cell按height降序排列，又可以怎么写？
compareTo函数中可以定义成这样：
return o.height - this.height;
这就意味着如果this.height > o.height时，Comparable返回的是当前元素“小于”o，又因为JAVA的PriorityQueue是小顶堆，因此队列的当前元素排在o之前，也就是队列是按降序排列的。
代码：
1 public class Solution {
2 class Cell implements Comparable<Cell> {
3 int row, col, height;
4 public Cell(int row, int col, int height) {
5 this.row = row;
6 this.col = col;
7 this.height = height;
8 }
9 //返回负数意味着当前元素“小于”o，返回正数意味着当前元素“大于”o，返回0意味着当前元素“等于”o
10 //JAVA中PriorityQueue默认的是小顶堆
11 public int compareTo(Cell o) {
12 return this.height - o.height;
13 }
14 }
15
16 public int trapRainWater(int[][] heightMap) {
17 int n = heightMap.length, m = heightMap[0].length, res = 0;
18 int[][] dirs = new int[][]{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
19 boolean[][] visited = new boolean[n][m];
20 PriorityQueue<Cell> queue = new PriorityQueue<Cell>();
21 for (int i = 0; i < n; ++i) {
22 visited[i][0] = visited[i][m - 1] = true;
23 queue.offer(new Cell(i, 0, heightMap[i][0]));
24 queue.offer(new Cell(i, m - 1, heightMap[i][m - 1]));
25 }
26 for (int i = 0; i < m; ++i) {
27 visited[0][i] = visited[n - 1][i] = true;
28 queue.offer(new Cell(0, i, heightMap[0][i]));
29 queue.offer(new Cell(n - 1, i, heightMap[n - 1][i]));
30 }
31 while (!queue.isEmpty()) {
32 Cell cell = queue.poll();
33 for (int i = 0; i < 4; ++i) {
34 for (int j = 0; j < 2; ++j) {
35 int row = cell.row + dirs[i][j];
36 int col = cell.col + dirs[i][j];
37 if (row > 0 && row < n && col > 0 && col < m && !visited[row][col]) {
38 res += Math.max(0, cell.height - heightMap[row][col]);
39 queue.offer(new Cell(row, col, Math.max(cell.height, heightMap[row][col])));
40 }
41 }
42 }
43 }
44 return res;
45 }
46 }