leetcode223题 题解 翻译 C语言版 Python版

223. Rectangle Area

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.



Assume that the total area is never beyond the maximum possible value of int.
223.矩形面积
在2D平面上求两个矩形覆盖的总面积。
每个矩形由左下角点和右上角点标识。
假设所有区域都没有超过int范围的点坐标。

思路：我们只要能求出两矩形相交的面积，然后用两矩形的面积和减去相交的面积即可。两矩形相交有多种情况，无论哪种情况都要确定相交部分的两个角点。相交部分左下角点由两个矩形的左下角点确定，分别取横纵坐标的大值，右上角点由两个矩形的右上角点确定，分别取横纵坐标的小值。这样就能很容易得到相交部分的面积，从而计算出总覆盖面积。

int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
int ldx = A > E ? A : E;
int ldy = B > F ? B : F;
int rtx = C < G ? C : G;
int rty = D < H ? D : H;
int intersecArea = 0;
if (ldx >= rtx || ldy >= rty) intersecArea = 0;
else intersecArea = (rtx - ldx) * (rty - ldy);
return (C-A)*(D-B) + (G-E)*(H-F) - intersecArea;
}

class Solution(object):
def computeArea(self, A, B, C, D, E, F, G, H):
"""
:type A: int
:type B: int
:type C: int
:type D: int
:type E: int
:type F: int
:type G: int
:type H: int
:rtype: int
"""
ldx, ldy, rtx, rty = max(A,E), max(B,F), min(C,G), min(D,H)
intersecArea = 0
if ldx >= rtx or ldy >= rty: intersecArea = 0
else: intersecArea = (rtx - ldx) * (rty - ldy)
return (C-A)*(D-B) + (G-E)*(H-F) - intersecArea