leetcode 223. 矩形面积

Posted 2021-03-03 wz-archer

在二维平面上计算出两个由直线构成的矩形重叠后形成的总面积。

每个矩形由其左下顶点和右上顶点坐标表示，如图所示。

 

 

 

示例:

输入: -3, 0, 3, 4, 0, -1, 9, 2
输出: 45
说明: 假设矩形面积不会超出 int 的范围。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/rectangle-area

class Solution {
public:
    int computeArea(int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4) {
        if(x1>x2)swap(x1,x2);
        if(y1>y2)swap(y1,y2);
        if(x3>x4)swap(x3,x4);
        if(y3>y4)swap(y3,y4);
        int num=0;

        int x=0;
        if(x1<=x3&&x4<=x2)x=x4-x3;
        else if(x3<=x1&&x2<=x4)x=x2-x1;
        else if(x2<=x4&&x3<=x2)x=x2-x1+x4-x3-(x4-x1);
        else if(x3<=x1&&x1<=x4)x=x2-x1+x4-x3-(x2-x3);

        int y=0;
        if(y1<=y3&&y4<=y2)y=y4-y3;
        else if(y3<=y1&&y2<=y4)y=y2-y1;
        else if(y2<=y4&&y3<=y2)y=y2-y1+y4-y3-(y4-y1);
        else if(y3<=y1&&y1<=y4)y=y2-y1+y4-y3-(y2-y3);
        
        return (x2-x1)*(y2-y1)-x*y+(x4-x3)*(y4-y3);
    }
};