LeetCode #223 Rectangle Area 矩形面积

223 Rectangle Area 矩形面积

Description:
Find the total area covered by two rectilinear rectangles in a 2D plane.

Each rectangle is defined by its bottom left corner and top right corner as shown in the figure.

Example:

rectangle area

Input: A = -3, B = 0, C = 3, D = 4, E = 0, F = -1, G = 9, H = 2
Output: 45

Note:

Assume that the total area is never beyond the maximum possible value of int.

题目描述:
在二维平面上计算出两个由直线构成的矩形重叠后形成的总面积。

每个矩形由其左下顶点和右上顶点坐标表示，如图所示。

示例 :

矩形

输入: -3, 0, 3, 4, 0, -1, 9, 2
输出: 45

说明：
假设矩形面积不会超出 int 的范围。

思路:

两个矩形面积之和减去两个矩形面积重叠的部分
时间复杂度O(1), 空间复杂度O(1)

代码:
C++:

class Solution 
{
public:
    int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) 
    {
        return (C - A) * (D - B) - (min(C, G) > max(A, E) ? min(C, G) - max(A, E) : 0) * (min(D, H) > max(B, F) ? min(D, H) - max(B, F) : 0) + (G - E) * (H - F);
    }
};

Java:

class Solution {
    public int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
        return (C - A) * (D - B) - (Math.min(C, G) > Math.max(A, E) ? Math.min(C, G) - Math.max(A, E) : 0) * (Math.min(D, H) > Math.max(B, F) ? Math.min(D, H) - Math.max(B, F) : 0) + (G - E) * (H - F);
    }
}

Python:

class Solution:
    def computeArea(self, A: int, B: int, C: int, D: int, E: int, F: int, G: int, H: int) -> int:
        return (C - A) * (D - B) + (G - E) * (H - F) - max(0, min(C, G) - max(A, E)) * max(0, min(D, H) - max(B, F))