Find the total area covered by two rectilinearrectangles in a 2D plane. Each rectangle is defined by its bottom left corner and top right corner as shown in the figure. Example: Input: A = -3, B = 0, C = 3, D = 4, E = 0, F = -1, G = 9, H = 2 Output:…
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. Cr…
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. 解法…
题目描述: 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 i…
翻译 找到在二维平面中两个相交矩形的总面积. 每一个矩形都定义了其左下角和右上角的坐标. (矩形例如以下图) 如果,总占地面积永远不会超过int的最大值. 原文 分析 这题前天试过,写了一堆推断.终究还是无果-- 贴几个别人的解决方式-- int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) { int64_t xmin1 = min( A, C ); int64_t xmax1 = max( A, C )…
/* 像是一道数据分析题 思路就是两个矩形面积之和减去叠加面积之和 */ public int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) { //求两个面积 int a1 = (C-A)*(D-B); int a2 = (G-E)*(H-F); //求叠加面积,(低上限-高下限)*(左右线-右左线) int h1 = Math.min(D,H); int h2 = Math.max(B,F); int…
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. Example: Input: A = -3, B = 0, C = 3, D = 4, E = 0, F = -1, G = 9…
We are given a list of (axis-aligned) rectangles. Each rectangle[i] = [x1, y1, x2, y2] , where (x1, y1) are the coordinates of the bottom-left corner, and (x2, y2) are the coordinates of the top-right corner of the ith rectangle. Find the total area…
