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【Lintcode】364.Trapping Rain Water II

2024-10-10 09:11:02   211人阅读

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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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 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;    
    }
};

 

【Lintcode】364.Trapping Rain Water II


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