> For the complete documentation index, see [llms.txt](https://nataliekung.gitbook.io/ladder_code/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://nataliekung.gitbook.io/ladder_code/qiang-hua-3-data-structure-ii/trapping-rain-water-ii.md).

# Trapping Rain Water II

Given*n\_x\_m\_non-negative integers representing an elevation map 2d where the area of each cell is\_1\_x\_1*, compute how much water it is able to trap after raining.

![](https://lintcode-media.s3.amazonaws.com/problem/trapping-rain-water-ii.jpg)

分析：

从外围最低点向内灌水 外面形成墙，保证灌水高度不会漏

1 最外围都加入堆，从外往内灌。

2 用堆维护最小值（override compare） 堆里加入三元组（x,y,当前灌水最高高度）

时间复杂度log(2m+2n)\*n\*m

农村包围城市，无需双重for loop 依然遍历了所有点，这种从外围然后四点扩展法值得注意。

答案：

```
 class Cell{
    public int x, y, h;
    Cell(int xx, int yy, int hh){
        x = xx;
        y = yy;
        h = hh;
    }
}
public class Solution{
    int[] dx = {1, -1, 0, 0};
    int[] dy = {0, 0, 1, -1};

    public int trapRainWater(int[][] heights){
        if(heights.length == 0)
            return 0;

        PriorityQueue<Cell> q = new PriorityQueue<Cell>(1, new Comparator<Cell>(){
            public int compare(Cell x, Cell y){
                return x.h - y.h;
            }
        });
        int n = heights.length;
        int m = heights[0].length;
        int[][] visit = new int[n][m];
        //初始化
        for(int i = 0; i < n; i ++){
            q.offer(new Cell(i, 0, heights[i][0]));
            q.offer(new Cell(i, m-1, heights[i][m-1]));
            visit[i][0] = 1;
            visit[i][m-1] = 1;
        }
        for(int i = 0; i < m; i ++){
            q.offer(new Cell(0, i, heights[0][i]));
            q.offer(new Cell(n-1, i, heights[n-1][i]));
            visit[0][i] = 1;
            visit[n-1][i] = 1;
        }
        int ans = 0;
        while(!q.isEmpty()){
            Cell now = q.poll();
            for(int i = 0; i < 4; i ++){
                int nx = now.x + dx[i];
                int ny = now.y + dy[i];
                if(0 <= nx && nx < n && 0 <= ny && ny < m && visit[nx][ny] == 0){
                    visit[nx][ny] = 1;
                    q.offer(new Cell(nx, ny, Math.max(now.h, heights[nx][ny])));
                    ans += Math.max(0, now.h - heights[nx][ny]);
                }
            }
        }
         return ans;
    }

}
```

Python

```
class Solution:
    def trapRainWater(self, heightMap: List[List[int]]) -> int:
        if not heightMap or not heightMap[0]:
            return 0
        n,m = len(heightMap),len(heightMap[0])
        q = [(heightMap[0][j],0,j) for j in range(m)] + [(heightMap[i][0],i,0) for i in range(n)] + [(heightMap[n-1][j],n-1,j) for j in range(m)] + [(heightMap[i][m-1],i,m-1) for i in range(n)]
        seen = {(0,j) for j in range(m)} | {(x,0) for x in range(n)}| {(n-1,j) for j in range(m)} | {(x,m-1) for x in range(n)}
        heapq.heapify(q)
        d = [-1,0,1,0,-1]
        res = 0
        while q:
            h,x,y = heapq.heappop(q)
            for nx,ny in [(x+d[i],y+d[i+1]) for i in range(4)]:
                if 0 <= nx < n and 0 <= ny < m and (nx,ny) not in seen:
                    if heightMap[nx][ny] < h:
                        res += h - heightMap[nx][ny]
                    heapq.heappush(q,(max(h, heightMap[nx][ny]),nx,ny))
                    seen.add((nx,ny))
        return res
```