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.

Example

Given 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 。

Solution 1 () (from 转自Granyang)

class Solution {public:    int trapRainWater(vector
     
      
        > &heightMap) {        if (heightMap.empty()) {            return 0;        }        int m = heightMap.size(), n = heightMap[0].size();        int res = 0, mx = INT_MIN;        priority_queue
       
        
         , vector
         
          
           >,greater
           
            
             >> q; vector
             
              
               > visited(m, vector
               
                (n, false)); vector
                
                 
                  > 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; }};