Given anm x nmatrix 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:
Bothmandnare 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.
2. Implementation
(1) BFS
public class Solution {
class Cell implements Comparable<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 compareTo(Cell that) {
return this.height - that.height;
}
}
public int trapRainWater(int[][] heightMap) {
if (heightMap == null || heightMap.length == 0 || heightMap[0].length == 0) {
return 0;
}
int m = heightMap.length, n = heightMap[0].length;
boolean[][] visited = new boolean[m][n];
PriorityQueue<Cell> queue = new PriorityQueue<>();
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || i == m - 1 || j == 0 || j == n - 1) {
queue.add(new Cell(i, j, heightMap[i][j]));
visited[i][j] = true;
}
}
}
int[][] directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
int curMaxHeight = 0;
int res = 0;
while (!queue.isEmpty()) {
Cell curCell = queue.remove();
curMaxHeight = Math.max(curMaxHeight, curCell.height);
for (int[] direction : directions) {
int nextRow = curCell.row + direction[0];
int nextCol = curCell.col + direction[1];
if (isValid(heightMap, nextRow, nextCol, visited)) {
visited[nextRow][nextCol] = true;
queue.add(new Cell(nextRow, nextCol, heightMap[nextRow][nextCol]));
if (heightMap[nextRow][nextCol] < curMaxHeight) {
res += curMaxHeight - heightMap[nextRow][nextCol];
}
}
}
}
return res;
}
public boolean isValid(int[][] heightMap, int row, int col, boolean[][] visited) {
return row >= 0 && row < heightMap.length && col >= 0 && col < heightMap[0].length && !visited[row][col];
}
}
3. Time & Space Complexity
BFS: 时间复杂度O(mn), 空间复杂度O(mn)
After the rain, water are trapped between the blocks. The total volume of water trapped is 4.
