Last Updated on 2022-11-21 by Clay
題目
Given the coordinates of two rectilinear rectangles in a 2D plane, return the total area covered by the two rectangles.
The first rectangle is defined by its bottom-left corner (ax1, ay1) and its top-right corner (ax2, ay2).
The second rectangle is defined by its bottom-left corner (bx1, by1) and its top-right corner (bx2, by2).
Example 1:
Input: ax1 = -3, ay1 = 0, ax2 = 3, ay2 = 4, bx1 = 0, by1 = -1, bx2 = 9, by2 = 2 Output: 45
Example 2:
Input: ax1 = -2, ay1 = -2, ax2 = 2, ay2 = 2, bx1 = -2, by1 = -2, bx2 = 2, by2 = 2 Output: 16
Constraints:
-104 <= ax1 <= ax2 <= 104-104 <= ay1 <= ay2 <= 104-104 <= bx1 <= bx2 <= 104-104 <= by1 <= by2 <= 104
本題其實非常單純，就是在計算兩個面積合併計算總共有多少。當然，兩個面積是可能存在著重疊（overlap）的情況。
解題紀錄
由於可能存在著重疊的面積，所以無論是在 X 軸還是 Y 軸上，我們都會想要取得『中間的兩個位置』。但是也可能存在著不重疊的情況，所以得額外判斷一下。
C++ 範例程式碼
class Solution {
public:
int computeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2) {
int aArea = (ax2-ax1)*(ay2-ay1);
int bArea = (bx2-bx1)*(by2-by1);
int xOverlap = max(min(ax2, bx2)-max(ax1, bx1), 0);
int yOverlap = max(min(ay2, by2)-max(ay1, by1), 0);
return aArea + bArea - xOverlap*yOverlap;
}
};Python 範例程式碼
class Solution:
def computeArea(self, ax1: int, ay1: int, ax2: int, ay2: int, bx1: int, by1: int, bx2: int, by2: int) -> int:
a_area = (ax2-ax1) * (ay2-ay1)
b_area = (bx2-bx1) * (by2-by1)
x_overlap = max(min(ax2, bx2)-max(ax1, bx1), 0);
y_overlap = max(min(ay2, by2)-max(ay1, by1), 0);
return a_area + b_area - x_overlap*y_overlap