leetcode 223. Rectangle Area 矩形面积计算

淩亂°似流年 2022-06-08 11:59 184阅读 0赞

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.

![这里写图片描述][SouthEast]

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;
        }
    };

[SouthEast]: /images/20220608/a23782cf513d4e82a7e2d768b88a9f01.png