[Leetcode] 407. Trapping Rain Water II 解题报告

题目：

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.

思路：

我记得Leetcode中好像有Trapping Rain Water I这道题目，我也贴出来解题报告了。其实这道题目就是将一维的情况扩展到了二维上，但是基本思路还是一样的：在一维中我们只需要从两个端点中选一个较低者即可，而在二维中可选的点就扩大成了整个矩形的四个边。根据上一道题目知道，我们每次应该先选取边界最小的高度，所以很自然的可以想到优先队列或者小顶堆来保存周围边界。在我们访问过了一个点之后，要继续在矩形的内部遍历，所以还需要保存一个点的位置信息（注意下面的代码将二维位置信息映射到了一维上）。为了防止再次访问已经访问过的点，还需要用一个数组来标记每个点的访问状态。

由于优先队列中的元素个数可达m*n个，所以该算法的时间复杂度应该是O(m*n*log(m*n))，空间复杂度是O(m*n)。

代码：

class Solution {public:    int trapRainWater(vector<vector<int>>& heightMap) {        if(heightMap.size() == 0) {            return 0;         }        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> que;    // smallest height in front        int row = heightMap.size(), col = heightMap[0].size();          vector<vector<int>> visited(row, vector<int>(col, 0));        int ans = 0, Max = INT_MIN;          for(int i = 0; i < row; i++) {              for(int j = 0; j < col; j++) {                  if(!(i == 0 || i == row - 1 || j == 0 || j == col - 1)) {   // boundary grid                    continue;                 }                que.push(make_pair(heightMap[i][j], i * col + j));          // put all the bounary grid in the que                visited[i][j] = 1;            }          }          vector<vector<int>> dir{{1, 0}, {-1, 0}, {0, 1}, {0, -1}};          while(!que.empty()) {              auto val = que.top();             que.pop();              int height = val.first, x = val.second / col, y = val.second % col;            Max = max(Max, height);                 for(auto d: dir) {      // for each direction                int x2 = x + d[0], y2 = y + d[1];                  if(x2 >= row || x2 < 0 || y2 < 0 || y2 >= col || visited[x2][y2]) {                    continue;                }                visited[x2][y2] = 1;                  if(heightMap[x2][y2] < Max) {                    ans += (Max - heightMap[x2][y2]);                  }                que.push(make_pair(heightMap[x2][y2], x2 * col + y2));            }          }          return ans;      }};