leetcode 407. Trapping Rain Water II

相关问题

11. Container With Most Water
42. Trapping Rain Water
407. Trapping Rain Water II

Discription

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

BFS
想象一下，随着持续降雨，水平面不断上升。随着水位上升，外界的水逐渐淹没小岛边缘，并填满低洼处，最后直至淹没小岛的最高处。

具体通过最小栈实现。
首先将小岛的边缘加入最小栈，如下图红色部分。假设此时水位为 ϵ > 0, 则所有边缘的土地都将被打湿（与外界水直接接触）。

取出并丢弃栈顶元素（下图绿色部分），不妨设其高度为h，当水面上升至 h+ϵ 时，开始从当前位置浸湿相邻干涩的土地。

若相邻干涩土地高于 h+ϵ， 则加入最小堆。

继续从栈顶取出并丢弃栈顶元素（下图绿色部分），水位上升

若相邻干涩土地低于此时水位h+ϵ， 则水将倒灌直至填满h+ϵ，如下图蓝色部分。将该土地加入最小堆（高度为h）。

重复以上步骤直至淹没（栈为空）。

以上图片来自Grandyang的博客

时间复杂度：?
空间复杂度：?

代码

typedef struct unit {    int r, c, h;    unit(int _r, int _c, int _h) : r(_r), c(_c), h(_h) {}}unit;class cmp {public:    bool operator()(const unit &a, const unit &b)    {        return a.h > b.h;    }};class Solution {public:    int trapRainWater(vector<vector<int>>& heightMap) {        if (heightMap.size() <= 1 || heightMap[0].size() <= 1)            return 0;        int rows = heightMap.size(), cols = heightMap[0].size();        vector<vector<bool>> visited(rows, vector<bool>(cols, false));        priority_queue<unit, vector<unit>, cmp> que;        vector<pair<int, int>> directions = { {0, 1}, {0, -1}, {1, 0}, {-1, 0} };        for (int i = 0; i < rows; i++)        {            visited[i][0] = true;            visited[i][cols - 1] = true;            que.push(unit(i, 0, heightMap[i][0]));            que.push(unit(i, cols - 1, heightMap[i][cols - 1]));        }        for (int j = 1; j < cols - 1; j++)        {            visited[0][j] = true;            visited[rows - 1][j] = true;            que.push(unit(0, j, heightMap[0][j]));            que.push(unit(rows - 1, j, heightMap[rows - 1][j]));        }        int sum = 0;        while (!que.empty())        {            unit tmp = que.top();                   que.pop();            for (int i = 0; i < directions.size(); i++)            {                int r = tmp.r + directions[i].first;                int c = tmp.c + directions[i].second;                if (0 <= r && r < rows && 0 <= c && c < cols && !visited[r][c])                {                    int h = heightMap[r][c];                    visited[r][c] = true;                    if (h < tmp.h)          //灌水                    {                        sum += tmp.h - h;                        h = tmp.h;                    }                    que.push(unit(r, c, h));                }            }        }        return sum;    }};