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 is trapped between the blocks. The total volume of water trapped is 4.

Approach #1: C++. [priority_queue]

class Solution {

public:

    int trapRainWater(vector<vector<int>>& heightMap) {

        if (heightMap.size() == 0) return 0;

        int row = heightMap.size(), col = heightMap[0].size();

        vector<vector<int>> visited(row, vector<int>(col, 0));

        priority_queue<pair<int, pair<int, int>>, vector<pair<int, pair<int, int>>>, greater<pair<int, pair<int, int>>>> pq;

        for (int i = 0; i < col; ++i) {

            pq.push({heightMap[0][i], {0, i}});

            pq.push({heightMap[row-1][i], {row-1, i}});

            visited[0][i] = 1;

            visited[row-1][i] = 1;

        }

        for (int i = 1; i < row-1; ++i) {

            pq.push({heightMap[i][0], {i, 0}});

            pq.push({heightMap[i][col-1], {i, col-1}});

            visited[i][0] = 1;

            visited[i][col-1] = 1;

        }

        int ans = 0;

        int curMaxHeight = 0;

        while (!pq.empty()) {

            pair<int, pair<int, int>> cur = pq.top();

            pq.pop();

            curMaxHeight = max(curMaxHeight, cur.first);

            int x = cur.second.first, y = cur.second.second;

            for (auto dir : dirs) {

                int xx = x + dir.first;

                int yy = y + dir.second;

                if (judge(xx, yy, heightMap) && visited[xx][yy] == 0) {

                    pq.push({heightMap[xx][yy], {xx, yy}});

                    visited[xx][yy] = 1;

                    if (heightMap[xx][yy] < curMaxHeight) {

                        ans += curMaxHeight - heightMap[xx][yy];

                    }

                }

            }

        }

        return ans;

    }

private:

    vector<pair<int, int>> dirs = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};

    static bool judge(int x, int y, vector<vector<int>>& heightMap) {

        int m = heightMap.size();

        int n = heightMap[0].size();

        if (x < 0 || x >= m || y < 0 || y >= n) return false;

        else return true;

    }

};

Analysis:

The problem is very typical of this similar questions.

Firstly, we use a priority_queue to store the bordars cells.

Secondly, we access to the top-first elements and record the maximum height in the top-first elements from start to now.

Thirdly, traveling current top-first element's top, left, right and bottom cells, if the position if vaild and the cell's height is less then the maximum height, then we use maximum height to subtract the cell's value, and add the difference to the ans.

巴特西

407. Trapping Rain Water II

最新文章

热门文章