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.
本题是一道简单的数学题，意思就是要求两个矩形的覆盖面积。
根据每个矩形是由它的下左边角和它的上右边角定义的特征，再结合公式：覆盖面积=两个矩形的面积-相交的面积，即可。
代码如下：
class Solution {
public:
int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
int area = (C-A)*(D-B) + (G-E)*(H-F);
if (A >= G || B >= H || C <= E || D <= F)
{
return area;
}
int top = (D>H)?H:D; //和用min(D,H)是一样的
int bottom = max(B, F);
int left = max(A, E);
int right = min(C, G);
return area - (top-bottom)*(right-left);
}
};
看了看别人做的，
class Solution {
public:
int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
if(A > E) return computeArea(E, F, G, H, A, B, C, D);
int res = (C - A)*(D- B) + (G - E)*(H - F);
if(C > E && B < H && F < D) res -= (min(C, G) - E) * (min(D, H) - max(B, F));
return res;
}
};
核心思想都是差不多的。
原文：http://www.cnblogs.com/carsonzhu/p/4568978.html