Leetcode 407. Trapping Rain Water II

Problem:

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.

Solution:

　　终于来到了赫赫有名的Trapping Rain Water II，和Leetcode 42. Trapping Rain Water完全不同，这道题使用了优先级队列加BFS，难度非常高，反正我是做不出来，这篇博客讲的非常详细了，我就引用他的博客吧。当时我想了一种效率非常低下的方法，仅供思维拓展，就是从顶部开始，每次取一层，小于该高度全部设为0，大于等于该高度设为1，计算这个01矩阵所有低凹处的体积，直到最底层为止，这样做有一个很大的缺陷，如果矩阵的最高值和最小值相差极大，效率就非常低。

Code:

 

class Solution {
public:
    int trapRainWater(vector<vector<int>>& heightMap) {
        int m = heightMap.size();
        if(m == 0) return 0;
        int n = heightMap[0].size();
        if(n == 0) return 0;
        priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>> pq;
        vector<vector<bool>> visited(m,vector<bool>(n,false));
        for(int i = 0;i < m;++i){
            visited[i][0] = true;
            visited[i][n-1] = true;
            pq.push(make_pair(heightMap[i][0],i*n));
            pq.push(make_pair(heightMap[i][n-1],i*n+n-1));
        }
        for(int j = 1;j < n-1;++j){
            visited[0][j] = true;
            visited[m-1][j] = true;
            pq.push(make_pair(heightMap[0][j],j));
            pq.push(make_pair(heightMap[m-1][j],(m-1)*n+j));
        }
        int result = 0;
        int height = 0;
        while(!pq.empty()){
            height = max(pq.top().first,height);
            int x = pq.top().second/n;
            int y = pq.top().second%n;
            pq.pop();
            if(x-1 >= 0 && !visited[x-1][y]){
                pq.push(make_pair(heightMap[x-1][y],(x-1)*n+y));
                visited[x-1][y] = true;
                result += ((height-heightMap[x-1][y]) > 0) ? (height-heightMap[x-1][y]) : 0;
            }
            if(x+1 < m && !visited[x+1][y]){
                pq.push(make_pair(heightMap[x+1][y],(x+1)*n+y));
                visited[x+1][y] = true;
                result += ((height-heightMap[x+1][y]) > 0) ? (height-heightMap[x+1][y]) : 0;
            }
            if(y-1 >= 0 && !visited[x][y-1]){
                pq.push(make_pair(heightMap[x][y-1],x*n+y-1));
                visited[x][y-1] = true;
                result += ((height-heightMap[x][y-1]) > 0) ? (height-heightMap[x][y-1]) : 0;
            }
            if(y+1 < n && !visited[x][y+1]){
                pq.push(make_pair(heightMap[x][y+1],x*n+y+1));
                visited[x][y+1] = true;
                result += ((height-heightMap[x][y+1]) > 0) ? (height-heightMap[x][y+1]) : 0;
            }
        }
        return result;
    }
};