leetcode 407 Trapping Rain Water II
z
 2019-04-17 11:36:39  

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:

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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.

img

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.

img

After the rain, water is trapped between the blocks. The total volume of water trapped is 4.

  • 我们从matrix的四周出发，一次将四周的点都保存到一个qriorityqueue中
  • 每次从queue中取出height最小的点t，从t出发，遍历他的四周的点s，如果s<t，说明s肯定能够trap住water的，而且，由于当前t是四周最小的点，所以s能够trap住(s-t)的water
  • 更新t中的height，将t加入pq中，直到所有的点都遍历完
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class Solution {
    class Cell {
        int x;
        int y;
        int val;
        public Cell(int x, int y, int val) {
            this.x = x;
            this.y = y;
            this.val = val;
        }
    }
    
    public int trapRainWater(int[][] heightMap) {
        if (heightMap == null || heightMap.length == 0 || heightMap[0].length == 0)
            return 0;
        PriorityQueue<Cell> pq = new PriorityQueue(1, new Comparator<Cell>(){
            public int compare(Cell a, Cell b) {
                return a.val - b.val;
            }
        });
        int water = 0;
        int m = heightMap.length;
        int n = heightMap[0].length;
        boolean[][] visit = new boolean[m][n];
        int[][] directions = new int[][]{{0, 1}, {0, -1}, {-1, 0}, {1, 0}};
        // 处理四周的cell
        for (int i=0; i<m; i++) {
            visit[i][0] = true;
            visit[i][n-1] = true;
            pq.offer(new Cell(i, 0, heightMap[i][0]));
            pq.offer(new Cell(i, n-1, heightMap[i][n-1]));
        }
        
        for (int i=0; i<n; i++) {
            visit[0][i] = true;
            visit[m-1][i] = true;
            pq.offer(new Cell(0, i, heightMap[0][i]));
            pq.offer(new Cell(m-1, i, heightMap[m-1][i]));
        }
        
        while (! pq.isEmpty()) {
            Cell t = pq.poll();
            for (int[] d: directions) {
                int x = d[0] + t.x;
                int y = d[1] + t.y;
                if (x>=0 && x < m && y>=0 && y<n && !visit[x][y]) {
                    visit[x][y] = true;
                    water += Math.max(0, t.val - heightMap[x][y]);
                    pq.offer(new Cell(x, y, Math.max(t.val, heightMap[x][y])));
                }
            }
        }
        return water;
    }
}