leetcode 223. Rectangle Area 矩形面积计算

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.

本题就是求解两个矩形的面积，注意要减去重叠面积。

这道题计算简单的面积计算问题，不涉及到太复杂的操作。

代码如下：

/* * 这个题就是简单的计算面积,不要想太多 * 要注意使用long类型 * 题目说坐标不超过int表示范围，但是计算举行边长可能超过int表示范围，所以这一点要注意 * */public class Solution {    public int computeArea(int A, int B, int C, int D, int E, int F, int G, int H)     {        long area1=(C-A)*(D-B);        long area2=(G-E)*(H-F);             long overLapArea=getOverLapArea(A,B,C,D,E,F,G,H);        return (int)(area1+area2-overLapArea);    }    public long getOverLapArea(int a, int b, int c, int d, int e, int f, int g,int h)     {               long aa=Math.max(a, e) , bb=Math.min(c, g);        long len1=(long)(bb-aa);        if(len1<=0)            return 0;        long cc=Math.max(b, f) , dd=Math.min(d, h);        long len2=(long)(dd-cc);        if(len2<=0)            return 0;        else             return len1*len2;    }}

下面是C++的做法，就是几个简单的计算问题

代码如下：

#include <iostream>#include <algorithm>#include <vector>#include <set>#include <map>using namespace std;class Solution {public:    int computeArea(int A, int B, int C, int D, int E, int F, int G, int H)    {        int area1 = (C - A)*(D - B);        int area2 = (G - E)*(H - F);        return area1 + area2 - get(A,B,C,D,E,F,G,H);    }    int get(int A, int B, int C, int D, int E, int F, int G, int H)    {        long aa = max(A, E), bb = min(C, G);        long len1 = bb - aa;        if (len1 <= 0)            return 0;        aa = max(B, F);        bb = min(D,H);        long len2 = bb - aa;        if (len2 <= 0)            return 0;        else            return len1*len2;    }};