leetcode223

 1 class Solution:
 2     def computeArea(self, A: int, B: int, C: int, D: int, E: int, F: int, G: int, H: int) -> int:
 3         x = 0
 4         y = 0
 5 
 6         if (A <= E):
 7             if (C <= E):
 8                 x = 0
 9             elif (C >= G):
10                 x = G - E
11             else:
12                 x = C - E
13         elif (G <= A):
14             x = 0
15         elif (G >= C):
16             x = C - A
17         else:
18             x = G - A
19 
20         if (B <= F):
21             if (D <= F):
22                 y = 0
23             elif (D >= H):
24                 y = H - F
25             else:
26                 y = D - F
27         elif (H <= B):
28             y = 0
29         elif (H >= D):
30             y = D - B
31         else:
32             y = H - B
33         
34         return (C - A) * (D - B) + (G - E) * (H - F) - x * y

算法思路，各种条件判断，画一幅图表示吧，对角的图形有重复的的，因此四个对角一共4种，其他位置一共10种，总共是14种情况。

posted on 2020-04-04 10:14 Sempron2800+ 阅读(133) 评论(0) 编辑收藏举报