Leetcode 42 407 Trapping Rain Water I II 双指针

最新推荐文章于 2023-01-15 19:29:43 发布

taoqick

最新推荐文章于 2023-01-15 19:29:43 发布

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本文介绍了一种方法，用于计算在给定的等高线地图上，雨后可以积聚的水量。此外，还探讨了如何计算二维矩阵中雨水的累积体积，通过遍历矩阵并应用特定规则来估算每个单元格的积水情况。

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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.

For example,
Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

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!

就是四面都是围墙，从最低的往里走；
如果里面有更低的，当然就可以蓄水，蓄水的量就是围墙最低减去此处的高度；
如果里面的比当前围墙高，那这个方向的围墙高度就增加了。
然后永远围墙最低的地方开始搜，最后就能把整个水池搜一遍

class Solution:
    def trap(self, height) -> int:
        l,r,wall,lower,res = 0,len(height)-1,0,0,0
        while (l<r):
            lower = min(height[l],height[r])
            res += max(0, wall-lower)
            (l,r) = (l+1,r) if height[l] <= height[r] else (l,r-1)
            wall = max(wall,lower)
        return res

s = Solution()
print(s.trap([0,1,0,2,1,0,1,3,2,1,2,1]))

--------------------------------------------------------

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.

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.

Constraints:

1 <= m, n <= 110
0 <= heightMap[i][j] <= 20000

------------------------------------------

import heapq
class Solution:
    def trapRainWater(self, heightMap) -> int:
        rows,cols = len(heightMap),len(heightMap[0]) if heightMap else 0
        if (rows == 0 or cols == 0):
            return 0
        pq,res,height = [],0,0
        for i in range(rows):
            for j in range(cols):
                if (i in {0,rows-1} or j in {0,cols-1}):
                    pq.append((heightMap[i][j],i,j))
                    heightMap[i][j] = -1
        heapq.heapify(pq)
        while (pq):
            ch,x,y = heapq.heappop(pq)
            height = max(ch, height)  # bug2: height = max(nh, height)
            for dx,dy in [(0,1),(1,0),(0,-1),(-1,0)]:
                nx,ny = x+dx,y+dy
                if (0<=nx<rows and 0<=ny<cols and heightMap[nx][ny] >= 0): #bug4: heightMap[nx][ny] > 0
                    nh = heightMap[nx][ny]
                    res += max(0, (height-nh))
                    heapq.heappush(pq,(nh,nx,ny))
                    heightMap[nx][ny] = -1 #bug1
                    #break #bug3: 周围是要保证都走一遍的
        return res