LeetCode Trapping Rain Water II

原题链接在这里：https://leetcode.com/problems/trapping-rain-water-ii/?tab=Description

题目：

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.

题解：

是Trapping Rain Water进阶题目.

从四周开始每次找边界中最小的值，这里利用minHeap, 然后向内扩展。

若最小边界高度大于当前高度，则求高度差为蓄水量，并把原高度放回minHeap中.

若最小边界高度小于当前高度，则改点不存水，把当前高度放回minHeap中.

Time Complexity: O(mn * log(m+n)).

m = heightMap.length, n = heightMap[0].length. 四周边界加进minHeap用时O((m+n) * log(m+n)). 中间部分用时O(m*n*log(m+n)), 每个点都有一次入minHeap和一次出minHeap.

Space: O(m*n). boolean array size.

AC Java:

1 public class Solution {
2     public int trapRainWater(int[][] heightMap) {
3         if(heightMap == null || heightMap.length < 3 || heightMap[0].length < 3){
4             return 0;
5         }
6         int m = heightMap.length;
7         int n = heightMap[0].length;
8         boolean [][] visited = new boolean[m]
;
9         PriorityQueue<Cell> minHeap = new PriorityQueue<Cell>(1, new Comparator<Cell>(){
10             public int compare(Cell c1, Cell c2){
11                 return c1.height - c2.height;
12             }
13         });
14
15         for(int i = 0; i<m; i++){
16             visited[i][0] = true;
17             visited[i][n-1] = true;
18             minHeap.offer(new Cell(i, 0, heightMap[i][0]));
19             minHeap.offer(new Cell(i, n-1, heightMap[i][n-1]));
20         }
21
22         for(int j = 0; j<n; j++){
23             visited[0][j] = true;
24             visited[m-1][j] = true;
25             minHeap.offer(new Cell(0, j, heightMap[0][j]));
26             minHeap.offer(new Cell(m-1, j, heightMap[m-1][j]));
27         }
28
29         int res = 0;
30         int [][] dirs = new int [][]{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
31         while(!minHeap.isEmpty()){
32             Cell c = minHeap.poll();
33             for(int [] dir : dirs){
34                 int x = c.i+dir[0];
35                 int y = c.j+dir[1];
36                 if(x>=0 && x<m && y>=0 && y<n && !visited[x][y]){
37                     visited[x][y] = true;
38                     res += Math.max(c.height, heightMap[x][y]) - heightMap[x][y];
39                     minHeap.offer(new Cell(x, y, Math.max(c.height, heightMap[x][y])));
40                 }
41             }
42         }
43         return res;
44     }
45 }
46
47 class Cell{
48     int i, j, height;
49     public Cell(int i, int j, int height){
50         this.i = i;
51         this.j = j;
52         this.height = height;
53     }
54 }