[leetcode]407. Trapping Rain Water II

xiaocong1990

已于 2024-05-01 16:09:13 修改

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于 2017-06-25 14:19:54 首次发布

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本文介绍了一个二维地图上计算雨水可捕捉体积的问题。通过使用优先队列遍历边界单元格并向内推进的方式，算法实现了有效计算。文章提供了两种实现方案，包括使用 pair 和 vector 的优先队列。

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题目链接：https://leetcode.com/problems/trapping-rain-water-ii/#/description

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.

class Solution{
public:
    int trapRainWater(vector<vector<int>>& heightMap) {
        if(heightMap.empty()) return 0;
        priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>> que;
        int row=heightMap.size(),col=heightMap[0].size();
        vector<vector<int>> visited(row,vector<int>(col,0));
        int ans=0,Max=INT32_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)) continue;
                que.push(make_pair(heightMap[i][j],i*col+j));
                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)
            {
                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;
    }
};

class Solution{
public:


struct cmp{
	bool operator()(vector<int> a, vector<int> b){
		return a[0]>b[0];
	}
};
    int trapRainWater(vector<vector<int>>& heightMap) {
        if(heightMap.empty()) return 0;
        priority_queue<vector<int>,vector<vector<int>>,cmp> que;
        int row=heightMap.size(),col=heightMap[0].size();
        vector<vector<int>> visited(row,vector<int>(col,0));
        int ans=0,Max=INT32_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)) continue;
                que.push({heightMap[i][j],i, j});
                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[0],x=val[1],y=val[2];
            Max=max(Max,height);
            for(auto d:dir)
            {
                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({heightMap[x2][y2],x2, y2});
            }
        }
        return ans;
    }
};