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.
Refer to https://discuss.leetcode.com/topic/60418/java-solution-using-priorityqueue/2
这里有具体的例子：http://www.cnblogs.com/grandyang/p/5928987.html
Analysis, 根据木桶原理，先找外围最矮的bar，里边如果有bar比它还矮，一定能存水（因为四周所有的bar都比它高）
注意还可能存更多的水，因为往里面，很可能cell高度变化。所以要把BFS中间遇到的高的bar都存进queue，随着水平面提升，提升到这些bar的高度，看能不能有凹槽存更多的水
44-45行逻辑就是
if (height[row][col] < cur) {
res += cur.height- height[row][col];
queue.offer(new Cell(row, col, cur.height));
}
else {
queue.offer(new Cell(row, col, height[row][col]));
}
public class Solution {
public class Cell {
int row;
int col;
int height;
public Cell(int x, int y, int val) {
this.row = x;
this.col = y;
this.height = val;
}
}
public int trapRainWater(int[][] heightMap) {
if (heightMap==null || heightMap.length<=2 || heightMap[0].length<=2) return 0;
int m = heightMap.length;
int n = heightMap[0].length;
int res = 0;
PriorityQueue<Cell> queue = new PriorityQueue<Cell>(1, new Comparator<Cell>() {
public int compare(Cell c1, Cell c2) {
return c1.height - c2.height;
}
});
HashSet<Integer> visited = new HashSet<Integer>();
for (int i=0; i<m; i++) {
queue.offer(new Cell(i, 0, heightMap[i][0]));
queue.offer(new Cell(i, n-1, heightMap[i][n-1]));
visited.add(i*n+0);
visited.add(i*n+n-1);
}
for (int j=0; j<n; j++) {
queue.offer(new Cell(0, j, heightMap[0][j]));
queue.offer(new Cell(m-1, j, heightMap[m-1][j]));
visited.add(0*n+j);
visited.add((m-1)*n+j);
}
int[][] directions = new int[][]{{-1, 0}, {1, 0}, {0, 1}, {0, -1}};
while (!queue.isEmpty()) {
Cell cur = queue.poll();
for (int[] dir : directions) {
int row = cur.row + dir[0];
int col = cur.col + dir[1];
if (row>=0 && row<m && col>=0 && col<n && !visited.contains(row*n+col)) {
visited.add(row*n+col);
res += Math.max(0, cur.height - heightMap[row][col]);
queue.offer(new Cell(row, col, Math.max(cur.height, heightMap[row][col])));
}
}
}
return res;
}
}