問題

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.

例子

-2 -2 2 2 -2 -2 2 2
16

分析

• 方法一：首先判斷兩個矩形是否相交。如果相交，總面積就是兩個矩形的面積和；如果不相交，總面積等于兩個矩形的面積和-重疊部分的面積；
• 方法二：通過min max函數直接求出重疊部分的面積，當矩形不相交時，重疊面積為0。

要點

分情況討論=>歸一化為一種情況

時間復雜度

O(1)

空間復雜度

O(1)

代碼

方法一

class Solution {
public:
    int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
        int total = (C - A) * (D - B) + (G - E) * (H - F);
        if (C <= E || G <= A || D <= F || H <= B) return total;
        vector<int> x{A, C, E, G};
        vector<int> y{B, D, F, H};
        sort(x.begin(), x.end());
        sort(y.begin(), y.end());
        int intersection = (x[2] - x[1]) * (y[2] - y[1]);
        return total - intersection;
    }
};

方法二

class Solution {
public:
    int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
        int left = max(A, E), right = max(min(C, G), left); // 當兩矩形不相交時，right = left
        int bottom = max(B, F), top = max(min(D, H), bottom); // 當兩矩形不相交時，top = bottom
        return (C - A) * (D - B) + (G - E) * (H - F) - (right - left) * (top - bottom);
    }
};