407. Trapping Rain Water II（python shixian）

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.

The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image!

Example:

Input: [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6      

def trapRainWater(self, heightMap):
        if not heightMap or not heightMap[0]:
            return 0
        
        import heapq    
        m, n = len(heightMap), len(heightMap[0])#m是行，n是列
        heap = []
        visited = [[0]*n for _ in range(m)]

        # Push all the block on the border into heap  这些都是初始化都带详解都好说啊！
        for i in range(m):
            for j in range(n):
                if i == 0 or j == 0 or i == m-1 or j == n-1:
                    heapq.heappush(heap, (heightMap[i][j], i, j))
                    visited[i][j] = 1
        
        result = 0
        while heap:
            height, i, j = heapq.heappop(heap)    #最小堆返回的是最小高度的值
            for x, y in ((i+1, j), (i-1, j), (i, j+1), (i, j-1)):
                if 0 <= x < m and 0 <= y < n and not visited[x][y]:
                    result += max(0, height-heightMap[x][y])
                    heapq.heappush(heap, (max(heightMap[x][y], height), x, y))#要加入这个最大值，水面的高度现在是这个高度了
                    visited[x][y] = 1
        return result