Lintcode364.Trapping Rain Water II

【Lintcode364.Trapping Rain Water II】青春须早为，岂能长少年。这篇文章主要讲述Lintcode364.Trapping Rain Water II相关的知识，希望能为你提供帮助。
题目：
Given n x m non-negative integers representing an elevation map 2d where the area of each cell is 1 x 1, compute how much water it is able to trap after raining.

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ExampleGiven 5*4 matrix

[12,13,0,12] [13,4,13,12] [13,8,10,12] [12,13,12,12] [13,13,13,13]

return 14.
题解：
之前的题是Two pointer，本质是在给定的边界不断向中心遍历，这道题也是，不过边界由两端变成了四个边，同样是内缩遍历。而且这道题还需要堆的思想来从最小端开始遍历（防止漏水）。详见 here。
Solution 1 () (from here 转自Granyang)

class Solution { public: int trapRainWater(vector< vector< int> > & heightMap) { if (heightMap.empty()) { return 0; } int m = heightMap.size(), n = heightMap[0].size(); int 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] == true) 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; } };