407. Trapping Rain Water II

407. Trapping Rain Water II

題目連接：
https://leetcode.com/problems...

參考discussion裏的解法：
https://discuss.leetcode.com/...

參考博客裏的解釋：
http://www.cnblogs.com/grandy...

public class Solution {
    public int trapRainWater(int[][] heightMap) {
        // base case
        if(heightMap.length == 0 || heightMap[0].length == 0) return 0;
        
        int m = heightMap.length, n = heightMap[0].length;
        // bfs, heap
        boolean[][] visited = new boolean[m][n];
        // add 4 sides first, since 4 side can not store water
        PriorityQueue<Cell> minHeap = new PriorityQueue<>((m+n), (a, b) -> a.h - b.h);
        // 1st col and last col
        for(int i = 0; i < m; i++) {
            minHeap.offer(new Cell(i, 0, heightMap[i][0]));
            visited[i][0] = true;
            minHeap.offer(new Cell(i, n - 1, heightMap[i][n-1]));
            visited[i][n-1] = true;
        }
        // 1st row and last row
        for(int j = 0; j < n; j++) {
            minHeap.offer(new Cell(0, j, heightMap[0][j]));
            visited[0][j] = true;
            minHeap.offer(new Cell(m-1, j, heightMap[m-1][j]));
            visited[m-1][j] = true;
        }
        // bfs find water
        int res = 0;
        while(!minHeap.isEmpty()) {
            Cell cur = minHeap.poll();
            for(int[] dir : dirs) {
                int nx = cur.x + dir[0], ny = cur.y + dir[1];
                if(nx >= 0 && nx < m && ny >= 0 && ny < n && !visited[nx][ny]) {
                    visited[nx][ny] = true;
                    if(heightMap[nx][ny] < cur.h) res += cur.h - heightMap[nx][ny];
                    minHeap.offer(new Cell(nx, ny, Math.max(cur.h, heightMap[nx][ny])));
                }
            }
        }
        return res;
    }
    
    int[][] dirs = new int[][] {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
}

class Cell {
    int x;
    int y;
    int h;
    Cell(int x, int y, int h) {
        this.x = x;
        this.y = y;
        this.h = h;
    }
}