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原题地址：https://leetcode-cn.com/problems/trapping-rain-water-ii/
题目描述：
给定一个 m x n 的矩阵，其中的值均为正整数，代表二维高度图每个单元的高度，请计算图中形状最多能接多少体积的雨水。
说明:
m 和 n 都是小于110的整数。每一个单位的高度都大于 0 且小于 20000。
示例：
给出如下 3x6 的高度图:
[
[1,4,3,1,3,2],
[3,2,1,3,2,4],
[2,3,3,2,3,1]
]
返回 4。
如上图所示，这是下雨前的高度图[[1,4,3,1,3,2],[3,2,1,3,2,4],[2,3,3,2,3,1]] 的状态。
下雨后，雨水将会被存储在这些方块中。总的接雨水量是4。
解题方案：
参考链接
木桶原理的衍生题，好难哦。。。
class Solution:
def trapRainWater(self, heightMap: List[List[int]]) -> int:
m = len(heightMap)
n = len(heightMap[0]) if m else 0
peakMap = [[float('inf')] * n for _ in range(m)]
queue = []
for x in range(m):
for y in range(n):
if x in (0, m - 1) or y in (0, n - 1):
peakMap[x][y] = heightMap[x][y]
queue.append((x, y))
while queue:
x, y = queue.pop(0)
for dx, dy in zip((1, 0, -1, 0), (0, 1, 0, -1)):
nx, ny = x + dx, y + dy
if nx <= 0 or nx >= m - 1 or ny <= 0 or ny >= n - 1:
continue
limit = max(peakMap[x][y], heightMap[nx][ny])
if peakMap[nx][ny] > limit:
peakMap[nx][ny] = limit
queue.append((nx, ny))
return sum(peakMap[x][y] - heightMap[x][y] for x in range(m) for y in range(n))
最佳：
class Solution:
def trapRainWater(self, heightMap: List[List[int]]) -> int:
if not heightMap or not heightMap[0]:
return 0
import heapq
m, n = len(heightMap), len(heightMap[0])
heap = []
visited = [[0]*n for _ in range(m)]
# Push all the block on the border into heap
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)
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))
visited[x][y] = 1
return result
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