Rectangle Area

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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.

解题思路：

简单计算几何。根据容斥原理：S(M ∪ N) = S(M) + S(N) - S(M ∩ N)

题目可以转化为计算矩形相交部分的面积

S(M) = (C - A) * (D - B)

S(N) = (G - E) * (H - F)

S(M ∩ N) = max(min(C, G) - max(A, E), 0) * max(min(D, H) - max(B, F), 0)

注意： min(C, G) - max(A, E)， min(D, H) - max(B, F)可能会溢出， 需要先转换成long 判断后再转换回int.

 Java code:

 1  public int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
 2         int sums = (C-A) *(D-B) + (G-E)*(H-F);
 3         long x = (long)Math.min(C,G)- (long)Math.max(A,E);
 4         long y = (long)Math.min(D,H)- (long)Math.max(B,F);
 5         if(x <  Integer.MIN_VALUE || x > Integer.MAX_VALUE) { x = 0;}
 6         if(y <  Integer.MIN_VALUE || y > Integer.MAX_VALUE) { y = 0;}
 7         int m = (int)x;
 8         int n = (int)y;
 9        return  sums - Math.max(m, 0) * Math.max(n, 0);
10     }

Reference:

1. http://bookshadow.com/weblog/2015/06/08/leetcode-rectangle-area/

Rectangle Area

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原文地址：http://www.cnblogs.com/anne-vista/p/4793469.html