# 407. Trapping Rain Water II

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.

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

Github 同步地址：

https://github.com/grandyang/leetcode/issues/407

Trapping Rain Water

https://leetcode.com/problems/trapping-rain-water-ii/

https://leetcode.com/problems/trapping-rain-water-ii/discuss/89461/Java-solution-using-PriorityQueue

https://leetcode.com/problems/trapping-rain-water-ii/discuss/89476/concise-C%2B%2B-priority_queue-solution

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