[Leetcode]-Rectangle Area

//题目: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.

//求两个正方形的面积(重叠地方只算一次)，已知两个正方形的对角顶点
//关键在于分别求取所有4个横纵轴坐标的两个中间值
//注意:当两个正方形无重叠的情况

#include <stdlib.h>#include <stdio.h>#define min(a,b) ((a)<(b))?(a):(b)#define max(a,b) ((a)>(b))?(a):(b)int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {    //正方形1对角顶点(A,B)     (C,D)      //正方形2对角顶点(E,F)     (G,H)    //横坐标: A,C,E,G    //纵坐标: B,D,F,H    //判断无重叠的情况    if((abs(A-G)-(C-A)-(G-E)) > 0)         return (G-E)*(H-F) + (D-B)*(C-A);    if((abs(E-C)-(C-A)-(G-E)) > 0)         return (G-E)*(H-F) + (D-B)*(C-A);    if((abs(H-B)-(D-B)-(H-F)) > 0)         return (G-E)*(H-F) + (D-B)*(C-A);    if((abs(D-F)-(D-B)-(H-F)) > 0)         return (G-E)*(H-F) + (D-B)*(C-A);    //if((A == B)&&(B == C)&&(C == D)) return (G-E)*(H-F);    //if((E == F)&&(F == G)&&(G == H)) return (D-B)*(C-A);    //有重叠的情况    int m=0,n=0;    n = abs((min(G,C))-(max(A,E)));    m = abs((min(D,H))-(max(B,F)));    return (G-E)*(H-F) + (D-B)*(C-A) - m*n;}int main(){    int A = -2,B = -2,C = 2,D = 2,E = -2,F = -2,G = 2,H = 2;    int area = computeArea( A, B, C, D, E, F, G, H);    printf("area is :%d\n",area);}