题目

题目

解题思路

解这道题的思路为，首先获取一个坐标点长方体的表面积，计算公式如下：

surface = grid[i][j] * 4 + 2

然后减去其 “上、下、左、右” 与相邻长方体的重叠表面积，减去的面积应该是相邻两个长方体中高度（grid[i][j]）较小的那个。

参考代码

代码一：为了方便理解，我们先写个不太精炼的, 如下：

'''
@auther: Jedi.L
@Date: Wed, Mar 27, 2019 12:44
@Email: xiangyangan@gmail.com
@Blog: www.tundrazone.com
'''

class Solution:
    def surfaceArea(self, grid):
        self.grid = grid
        tsa = 0  # total surface area of resulting shapes
        x = len(self.grid) - 1  # max i for grid[i][j]
        y = len((self.grid[0])) - 1  # max j for grid[i][j]
        i = 0
        while i <= x:
            j = 0
            while j <= y:
                if self.grid[i][j] != 0:
                    tsa += self.grid[i][j] * 4 + 2  # surface are at point
                    # we count the left surface area of this point as follow:
                    # up close
                    if i-1 >= 0 and self.grid[i-1][j] >= self.grid[i][j]:
                        tsa = tsa - self.grid[i][j]
                    elif i-1 >= 0 and self.grid[i-1][j] < self.grid[i][j]:
                        tsa = tsa - self.grid[i-1][j]
                    # down close
                    if i+1 <= x and self.grid[i+1][j] >= self.grid[i][j]:
                        tsa = tsa - self.grid[i][j]
                    elif i+1 <= x and self.grid[i+1][j] < self.grid[i][j]:
                        tsa = tsa - self.grid[i+1][j]
                    # left close
                    if j-1 >= 0 and self.grid[i][j-1] >= self.grid[i][j]:
                        tsa = tsa - self.grid[i][j]
                    elif j-1 >= 0 and self.grid[i][j-1] < self.grid[i][j]:
                        tsa = tsa - self.grid[i][j-1]
                    # right close
                    if j + 1 <= y and self.grid[i][j+1] >= self.grid[i][j]:
                        tsa = tsa - self.grid[i][j]
                    elif j + 1 <= y and self.grid[i][j+1] < self.grid[i][j]:
                        tsa = tsa - self.grid[i][j+1]
                j = j + 1
            i = i + 1
        return tsa

代码二：仔细观察代码一，我们会发现其中的一些代码没有必要，每个长方体只需要考虑自己的下方和右方相邻的长方体，然后减去2倍的重叠面积就可以了，精简后的代码如下：

'''
@auther: Jedi.L
@Date: Wed, Mar 27, 2019 12:44
@Email: xiangyangan@gmail.com
@Blog: www.tundrazone.com
'''

class Solution:
    def surfaceArea(self, grid):
        self.grid = grid
        tsa = 0  # total surface area of resulting shapes
        x = len(self.grid) - 1  # max i for grid[i][j]
        y = len((self.grid[0])) - 1  # max j for grid[i][j]
        i = 0
        while i <= x:
            j = 0
            while j <= y:
                if self.grid[i][j] != 0:
                    tsa += self.grid[i][j] * 4 + 2  # surface are at point
                    # we count the left surface area of this point as follow:
                    # down close
                    if i + 1 <= x and self.grid[i + 1][j] >= self.grid[i][j]:
                        tsa = tsa - 2*self.grid[i][j]
                    elif i + 1 <= x and self.grid[i + 1][j] < self.grid[i][j]:
                        tsa = tsa - 2*self.grid[i + 1][j]
                    # right close
                    if j + 1 <= y and self.grid[i][j + 1] >= self.grid[i][j]:
                        tsa = tsa - 2*self.grid[i][j]
                    elif j + 1 <= y and self.grid[i][j + 1] < self.grid[i][j]:
                        tsa = tsa - 2*self.grid[i][j + 1]
                j = j + 1
            i = i + 1
        return tsa

源码地址：
https://github.com/jediL/LeetCodeByPython

其它题目：[leetcode题目答案讲解汇总（Python版 持续更新）]
(https://www.jianshu.com/p/60b5241ca28e)

ps：如果您有好的建议，欢迎交流 :-D，
也欢迎访问我的个人博客 苔原带 (www.tundrazone.com)