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.

此题是Trapping Rain Water的变体，开始并没有想出来，看了大神的答案才恍然大悟。首先先思考再trapping rain water是怎么做的，我们用two pointer从外向里遍历，而此题方法与那道题目类似，是从外向里遍历，那么如何把边界包括进去呢？这里面需要用到小顶堆的用法了，首先定义一个cell类，存放row，col和height；再定义一个visited二维数组，用来确定是否访问过该cell；然后把边界的这些cell存放到小顶堆里面，因为小顶堆有自动按照从小到达将height排序的功能，先取出来的是较小height的cell，然后检测cell四周没有访问过的且不超过边界值的cell高度，如果未访问过的cell高度小于边界的高度，则高度差为该点存水的容积，否则不存水，全部遍历就是所要结果了，代码如下：

public class Solution {

    public class Cell{

        int row;

        int col;

        int height;

        public Cell(int row,int col,int height){

            this.row = row;

            this.col = col;

            this.height = height;

        }

    }

    public int trapRainWater(int[][] heightMap) {

        if(heightMap==null||heightMap.length==0||heightMap[0].length==0) return 0;

        int m = heightMap.length;

        int n = heightMap[0].length;

        boolean[][] visited =new boolean[m][n];

        int sum = 0;

        PriorityQueue<Cell> pq = new PriorityQueue<>(1,new Comparator<Cell>(){

            public int compare(Cell a,Cell b){

                return a.height-b.height;

            }

        });

        for(int i=0;i<m;i++){

            visited[i][0] = true;

            visited[i][n-1] = true;

            pq.offer(new Cell(i,0,heightMap[i][0]));

            pq.offer(new Cell(i,n-1,heightMap[i][n-1]));

        }

        for(int i=0;i<n;i++){

            visited[0][i]= true;

            visited[m-1][i] = true;

            pq.offer(new Cell(0,i,heightMap[0][i]));

            pq.offer(new Cell(m-1,i,heightMap[m-1][i]));

        }

        int[][] dirs = new int[][]{{0,1},{0,-1},{-1,0},{1,0}};

        while(!pq.isEmpty()){

            Cell cell = pq.poll();

            for(int[] dir:dirs){

                int row = cell.row+dir[0];

                int col = cell.col+dir[1];

                if(row>=0&&row<m&&col>=0&&col<n&&!visited[row][col]){

                    visited[row][col] = true;

                    sum+=Math.max(0,cell.height-heightMap[row][col]);

                    pq.offer(new Cell(row,col,Math.max(cell.height,heightMap[row][col])));

                }

            }

        }

        return sum;

    }

}

407. Trapping Rain Water II