LeetCode407. Trapping Rain Water II

最新推荐文章于 2021-11-04 09:56:06 发布

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本文介绍了一种计算二维高度地图中雨水可聚集体积的算法。通过分析单元格高度，利用优先队列和访问标记，从边界开始逐步填充内部区域，计算雨水体积。示例展示了3x6高度地图上雨水的聚集情况。

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

参考资料：here。

class Solution {
public:
	int trapRainWater(vector<vector<int>>& heightMap) {
		if (heightMap.empty()) return 0;
		int m = heightMap.size(), n = heightMap[0].size(), ans = 0, elevation = 0;
		vector<vector<int>> dirs{ {-1,0},{1,0},{0,-1},{0,1} };
		vector<vector<bool>> visited(m, vector<bool>(n, false));
		priority_queue < pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> q;
		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) {
					q.push({ heightMap[i][j],i * n + j });
					visited[i][j] = true;
				}
			}
		}
		while (!q.empty()) {
			pair<int, int> tmp = q.top(); q.pop();
			int h = tmp.first, r = tmp.second / n, c = tmp.second % n;
			elevation = max(elevation, h);
			for (vector<int> dir : dirs) {
				int x = r + dir[0], y = c + dir[1];
				if (x < 0 || x >= m || y < 0 || y >= n || visited[x][y]) continue;
				if (elevation > heightMap[x][y]) ans += elevation - heightMap[x][y];
				visited[x][y] = true;
				q.push({ heightMap[x][y],x * n + y });
			}
		}
		return ans;
	}
};